A kicked soccer ball has an initial velocity of 25 meters per second at an angle of 40.° above the horizontal,?
Well, you should really learn this on your own because you will need to know this on the test. But since you already know the awnser You did find the force in the y direction correctly, good job, that's probably the hardest part. First you have to find the time it takes to reach the maximum height. its derived from this equation v = v0 sin α − g t you kind of have to modify it to get to this t= vo * sin (a) / g you should then have (25 m / s * sin40)/9.81 m/s/s since your dividing m/s by m/s/s your answer should come out to be in seconds right? approximately 1.63976 seconds right? they you can use y=1/2g((1.63976)^2) and you should get 13.1752 assuming gravity = 9.81m/s^2
How to calculate initial velocity of a soccer ball?
If I just hit a ball and measured the distance within the first few meters, and also determined the time within that distance, would I have initial velocity? That's instantaneous velocity.. right? How do I get initial if I were doing it in real life?
A person kicks a soccer ball with an initial velocity directed 538 above the horizontal.physic problem of grade 12?
A person kicks a soccer ball with an initial velocity directed 538 above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25 m away, and it takes the ball 2.1 s to pass directly over the wall
A person kicks a soccer ball with an initial velocity directed 53 degrees above the horizontal. The ball lands?
Initial vertical velocity = v * sin 53, Initial horizontal velocity = v * cos 53 The vertical velocity decreases at the rate of 9.8 m/s each second. The horizontal velocity is constant. Use the following equation to determine the initial vertical velocity. d = vi * t + ½ * a * t^2 d represents the ball’s vertical displacement. Vertical displacement = Final height – Initial height d = 7.2 – 0 = 7.2 m vi = v * sin 53, a = -9.8 7.2 = v * sin 53 – 4.9 * 2.1^2 v * sin 53 = 28.809 This is the ball’s initial vertical velocity. v = 28.809 ÷ sin 53 The ball’s initial velocity is approximately 36.07 m/s. Use this velocity and the time in the following to determine the range. d = v * t, v = initial horizontal velocity = v * cos 53 d = 28.809 ÷ sin 53 * cos 53 * 2.1 = 61.1289 * ctn 53 This is approximately 45.6 meters. To determine the vertical distance the ball clears the wall, we need to determine the maximum height of the ball. As the ball moves to its maximum height, its vertical velocity decreases from the initial vertical velocity to 0 m/s. Use the following equation to determine the ball’s maximum height. vf^2 = vi^2 + 2 * a * d, vf = 0, vi = v * sin 53 = 28.809, a = -9.8 0 = 28.809^2 + 2 * -9.8 * d d = -28.809^2 ÷ -19.6 The maximum height is approximately 42.34 meters. To determine the vertical distance the ball clears the wall, subtract 7.2 meters. This is approximately 35.84 meters. I hope this helps you to understand how to solve this type of problem. If you have other questions that you need help in the future, make me one of your contacts. Your questions will come directly to the following yahoo email address. morrison60967@yah00.com
A ball is thrown upward with an initial velocity of 30 m/s. what is the maximum height that the ball will reach?
All answers I've read just give you an equation and fill it in. I just don't think that helps anyone understand what is being asked. So here goes my attempt at teaching anyone interested.There actually are several ways to calculate this of which I pick the one that is simplest to me.Assuming there is no (significant) air drag no energy is gained or lost; the kinetic energy (Ek) the ball has leaving your hand is converted in potential gravitational energy (Ep) on the way up, and back to kinetic energy on the way down.Their size must therefore be equal:Ek = EpAlso we know that,Ek = 0.5*m*v^2. and, Ep = m*g*hwith m the mass of the ball, v its velocity leaving your hand, g the gravitational acceleration and h the maximum height the ball reaches.Rearanging a bit gives you:v^2 = 2*g*h or,v^2/2g =hCurious about a different way?Take,St = S0 + V0 *t + 0.5*a*t^2which gives the position of an object at time t (St) given the position at time 0 (S0) the initial velocity (V0) and the acceleration (a) which is constant.For a we fill in -g (because down is negative) we take S0 = 0 and we know V0.To find t we need to solve it from,Vt = V0 + a*t.Since the velocity at the summit of the balls flight is zero we get:V0 = g*t. or, t = V0/g(Notice how I cleverly got rid of a minus?)Fill in that value into the equation for St and you're done.But you can also plug the formula for t into the equation:St = V0*V0/g - 0.5*g*(V0/g)^2= (V0^2)/g - 0.5*g*(V0^2)/g^2= (V0^2)/g - 0.5*(V0^2)/g= 0.5*(V0^2)/g = (V0^2)/2gwhich is the result we had before when you consider that v there is V0 here.Isn't it great how everything lines up?I hope this helped you understand rather than get the answer right.Yes… There's a difference…
A kicked soccer ball has an initial velocity of 25 meters per second at an angle of 40 degrees above the?
the initial vertical velocity of the ball is 25 sin 40 = 16.1m/s since gravity reduces the vertical velocity of objects by 9.8m/s/s, this ball will reach max altitude in a time of time = 16.1m/s / 9.8m/s/s = 1.6s the max height is equivalent to the height an object will fall in 1.6s, so we have height =1/2 g t^2 = 1/2 x 9.8m/s/s x (1.6s)^2 = 13.2m
A soccer ball is kicked from the ground with an initial speed of 14.0 m/s. after .275 seconds its speed of?
When the ball is kicked, it has some initial horizontal velocity and an initial vertical velocity. As the ball travels through the air, the horizontal velocity stays the same, while the vertical velocity is reduced due to gravity. Let the angle with the ground be θ. The initial horizontal velocity is 14 * cos θ. The initial vertical velocity is 14 * sin θ, and after .275 seconds it has been reduced by .275 * g. The square of the new velocity, 12.9, is the sum of the squares of the horizontal and vertical velocities (Pythagorean theorem). 12.9^2 = (14 * cos θ)^2 + (14 * sin θ - .275 * g)^2 Solve for θ. Your solution will simplify if you remember that sin^2 (θ) + cos^2 (θ) = 1.
A soccer ball is kicked with an initial speed of 10.4m/s in a direction 20.0 degrees above the horizontal.?
Identifying the vertical and horizontal components is the real secret to solving projectile motion problems. The vertical component involves using the equations below: v = u + at …….. OR ……. v(y) = u(y) + at v² = u² + 2as ….. OR …… v(y)² = u(y)² + 2as s = ut + ½at² ….. OR …… s(y) = u(y)t + ½at² (Note: all the equations above contain an acceleration term which will be a = g = -9.8 m/s²) The horizontal component involves this equation: v = s / t …. OR …. v(x) = s(x) / t (NB: no acceleration; just constant velocity) One last tip: positive (+) = upward negative (-) = downward --------------------------------------... Solution: Vertical component: v(y) = u(y) + at v(y) = 10.4*sin20° + (-9.8*0.35) v(y) = 3.56 – 3.43 = 0.13 m/s (moving up) Horizontal velocity: v(x) = 10.4 m/s (horizontal) The actual velocity is calculated by vectorially combining both the vertical and horizontal components, using geometry: v² = v(x)² + v(y)² v = √ (0.13² + 10.4²) = √ (0.017 + 108.2) v = 10.403 m/s Direction: Tan Φ = 0.13 / 10.4 Φ = 0.72° from horizontal, heading upward