How long does it take the ball to reach its highest point?
a) At its highest point the balls velocity is 0 m/s so we can use the following equation to solve for t Vf = Vi - gt 0 = 27 - 9.8 t t = 27 / 9.8 = 2.755 seconds b) How high does the ball rise? Let's use this equation to figure that out Vf^2 = Vi^2 - 2gh Where again Vf= final velocity = 0 m/s, Vi is initial velocity = 27 m/s , h is the distance height solve for h h = (Vf^2 - Vi^2 ) / 2g = (0 - 27^2) / (2 *9.8) = 37.2 m
A ball is thrown straight up at 28 m/s. How long does it take for the ball to be 6.7 m above its release point?
The height h at time t is given by:h=28*t-0.5*g*t^2=28*t-4.9*t^2If h = 6.7 m, we must solve:6.7 = 28*t-4.9*t^2This is a quadratic equation whose solutions are:t1 = 0.25st2 = 5.46sThe first solution, 0.25s after being thrown on its way up.The second solution, 5.46s is on the way down.Maximum height is reach at time t3 such that:28–gt3=0So: t3 =28/g =28/9.8 = 2.86 s
A ball is thrown straight up to the sky with an initial velocity of 30 m/s. How long does it take the ball to fall back to the earth?
Depends on initial height above ground.It will take 3.06 s for the ball’s velocity to reach zero and for the ball to reach its maximum height, which will be 138 m above the initial height. This takes into account that gravitational acceleration ([math]g[/math]) is [math]-9.8 m/s^2[/math], initial velocity ([math]v[/math]) is [math]30 m/s[/math] and distance is [math]vt + 1/2 gt^2[/math]If the initial height above ground is denoted by [math]h[/math], then the time taken for the ball to hit the ground would be 3.06 s required to reach maximum height plus the time taken to fall back down to the ground.Time taken for an object to fall from certain height is [math]\sqrt{2d/g}[/math], so substitute [math]d[/math] for [math]h + 138[/math] and add 3.1 s to find the time taken for the ball to fall to the ground.
Ball thrown up at 30m/s. how high does it go and how long is it in the air?
If a ball is thrown directly vertically upwards at 30 m/s then its kinetic energy at the start is: - E=0.5mv^2 When the ball reaches the top of its climb and stops climbing it will be at a height h with a gravitational potential energy, of: - P=mgh Since, all of the kinetic energy has been converted into gravitational potential energy, we have: - E=P or 0.5mv^2 = mgh Now, eliminating m and rearranging for h gives: - h=(0.5v^2)/g The maximum height is thus: - h=0.5x30x30/9.81 Hence, h=45.872 m (rounded to 3 DP) The ball's rise and fall take the same time because E=P and thus: - h = 0.5gt^2 Or rearranging for t: - t = sqrt(2h/g) Or t=sqrt(2 x 45.872/9.81) Hence, t=2t the total time that the ball is in the air is t=6.116 seconds (rounded to 3DP) I hope this helps!
A ball is thrown straight up with an initial speed of 30 m/s.?
We need to use one of the kinematic equations to answer this question. We must notice two things to help us solve the problem 1/. At the highest point, the vertical component of the velocity = 0 because it momentarily stops as it turns around from going upwards to going downwards 2/. Since we're ignoring air resistance, the trajectory of the ball is symmetric. This means the time taken on the upward leg of the ball's journey is equal to the time taken on the downward leg. This means that the total time the ball is in the air is double the time taken to reach the highest point. What information have we got to solve the equation? u = initial speed = 30 m/s^2 v = "final" speed = 0 a = accel due to gravity = -9.8 m/s^2 (negative because downward) d = distance = ? So to solve this problem we note we must use the equation v^2 = u^2 + 2ad 0 = 30^2 - 19.6d d = 900/19.6 = 46 m. That's how high it goes. Since we know v, u and a we can use v = u + at to find the height to the highest point (and then double it). v = u + at 0 = 30 - 9.8t t = 30/9.8 = 3.06 seconds and doubling gives 6.12 seconds = 6.1 s.
A rock is thrown straight up with an initial velocity of 15 m/s. How high will the rock rise?Ignore air fricti?
the first thing you always want to do is list the variables that you know we know the initial velocity - 15m/s we know gravity - 9.8m/s^2 we know the final velocity - 0m/s we don't know - distance to apogee (highest point) well there is one equation that uses all 4 of these variables, we know three of them, so we can solve the unknown "d" value the equation is V_final^2 = V_initial^2 + 2 * a *d V_final = 0 V_initial = 15 a = -9.8m/s^2 d = ? k plug some stuff in 0^2 = 15^2 + 2*(-9.8)*d 0 = 225 - 19.6*d 19.6*d = 225 d = 225/19.6 d = 11.47959184 This means that the rock that you throw into the air will go about 11.48m before stopping, and cominb back down. If you wanted to solve for the time of this trip, you could simply use the equation below V_final = V_intial + a*t 0 = 15 + (-9.8)*t t = 15/9.8 = 1.530612245 seconds it will take 1.53 seconds to reach the top, and then another 1.53 to come back down meaning a total trip of 3.061224490 seconds I hope this helped!
A Ball is thrown straight upward with a speed of +12 m/s?
Weight = mass * g Force = mass * a The weight of an object is the force causing it to accelerate toward the earth. mass * a = mass * g a = g = 9.8 m/s^2 The acceleration, “due to gravity”, is g. In these questions, the value of g = 10 m/s^2 So, As the ball moves upward, the vertical velocity of an object decreases 10 m/s each second. When the ball reaches its maximum height, the vertical velocity is 0 m/s. As the ball moves downward, the vertical velocity of an object increases 10 m/s each second. The path is symmetrical. The time up equals the time down. The final vertical velocity, after falling back down is equal to the initial vertical velocity. Time up = (change of vertical velocity) ÷ g = (12 – 0) ÷ 10 = 1.2 s Maximum height = Average velocity up * Time up Average velocity up = ½ * (12 + 0) = 6 Maximum height = 6 * 1.2 A Ball is thrown straight upward with a speed of +12 m/s? -What is the ball's acceleration just after it is thrown? (A) 10 m/s^2 upward (B) 10 m/s^2 downward (C) 12 m/s^2 upward (D 12 m/s^2 downward (E) -how much time does it take for the ball to rise to its maximum height? (A) 24 s (B) 12 s (C) 10 s (D) 2 s (E) 1.2 s -what is the approximate maximum height the ball reaches? (A) 24 m (B) 17 m (C) 12 m (D)7m (E) 5m Totally different question: A ball is dropped from rest. WHat is the acceleration of the ball immediately after it is dropped? The acceleration, “due to gravity”, is g. In these questions, the value of g = 10 m/s^2 (A) zero (B) 5 m/s^2 (C) 10 m/s^2 (D) 20 m/s^2 (E) 30 m/s^2
A ball is thrown straight up at 25 m/s above level ground. How long does it take to return to the ground?
So what you do here is first figure out how the ball travels from the beginning to its highest point where it stops. Gravity is 9.81. So: v = vi + at 0 = 25m/s + (-9.81m/s^2)t (v is equal to 0 because at the highest point it does not move) t = 2.55 s It takes the ball the same amount of time to come back down to the ground so just multiply the time x 2, Therefore 2.55 s x 2 = 5.10 s Alright so for the second part you need to use another equation. d = vit + 1/2at^2 d = (25 m/s)(2.55s) + (1/2)(-9.81m/s)(2.55^2) d = 31.86 Now you just times it by two because it goes up and then down so 31.86 x 2 = 63.72 m. I don't know what you have to round to so try and adjust your answers. Hope this helped. : D
A ball thrown vertically upward reaches a maximum height of 30.0 meters above the ground. How fast was the ball thrown?
Kinetic energy of the throw was turned into potential energy of the height gained.1/2 m v^2 = m g hWherem is the mass of the ball, which will cancel out, so its exact value is not needed.v is the initial velocity of the ball.g is acceleration of gravity near Earth, 9.81 m/s^2h is the max height reached by the ball, 30 m.1/2 m v^2 = m (9.81) (30)Divide both sides by m.1/2 v^2 = 294.3v^2 = 588.6v = 24.26 m/s
A ball is thrown vertically upward. For how long does the ball remain stationary at the highest point of its trajectory?
4.5.2017 — “A ball is thrown vertically upward. For how long does the ball remain stationary at the highest point of its trajectory?”My intuition is that the time is [math]zero[/math]. That might be a little counter-intuitive to some people because the velocity is zero at the highest point. So let’s justify my intuition.Regarding the top of the trajectory as the origin of a vertical coordinate [math]y[/math], consider the time taken to go from [math]y=-h[/math], through [math]y=0[/math], and back to [math]y=-h[/math]. I’ll assume constant acceleration [math]-g[/math] and, sparing the reader the details*, the result is [math]time=2\sqrt{\frac{2h}{g}}[/math].The time at the highest point is less than this time for any value of [math]h[/math], no matter how small.The only ‘length’ of time that satisfies this is [math]zero[/math].* Using the equation for constant acceleration, [math]s=ut+\frac{1}{2}a{t}^{2}[/math], where [math]s=distance[/math], [math]u=initial[/math] [math]velocity[/math], [math]a=acceleration[/math], and [math]t=time[/math].This is a nice question—the mathematics is simple, but it asks us to think about instantaneous states of motion. Edit—this theme recurs in a comment and my response.