A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft away from him horizontally. If he throws a packet directly aiming at friend with a speed of 15 ft/s, how short will the packet fall?
✔Since the person on the cliff throws the stone directly towrds the person on ground, then while throwing he makes an angle θ (say) with the vertical cliff.▶tanθ= 228/171 = 4/3▶θ= 53o✔Taking the standard x-y coordinate system,Iin 'y' direction,✔Inital velocity = 15 cos θ = 15 cos 53° = 9 ft/sv^2 =u^2 + 2gh where g=32 ft/s, h=171ft, u=9ft/sv = 105ft/su + gt => 105 = 9 + 32tt = 3 s✔Now in 'x' direction, initial velocity = 15 sin 53 = 12 ft/s✔Distance travelled in 3 s = 12 * 3 = 36 ft▶➡Distance through which it falls short = 228 - 36 =192 ft
A man throws a stone over a cliff and hears it strike at the bottom in 8 seconds. The temperature is 25°c, what are the two times, the time that the stone travels and the time that sound travels? What is the height of the cliff?
You can solve this with a system of equations. Here’s how to set it up:Call the time it takes for the stone to fall to the bottom of the cliff [math]t_1[/math], and call the time it takes for the sound to return [math]t_2[/math] . This gives us one of our equations:[math]t_1 + t_2 = 8\ \text{seconds}[/math]But we also know that the distance, [math]d[/math], the stone falls is given by:[math]d = \dfrac{g}{2}{t_1}^{2}[/math]where [math]g[/math] is the acceleration rate of gravity.The distance can also be written in terms of the speed of sound, [math]s[/math]:[math]d = st_2[/math]where [math]s[/math] is temperature dependent (this is where the 25 C comes in).I’ll bet you can solve it from here.
A man standing on the top of a tower throws a ball straight up with some velocity, and at the same time throws a second ball straight downward with the same speed. Which ball will have a greater speed as they strike the ground?
If the tower is high enough, say a mile, and is placed on Planet Earth, both balls will strike the ground (that assumes you are throwing them outside the tower’s perimeter) at the same terminal speed. I am assuming both balls are the same diameter and are of same density. BTW, if you were doing this on the Moon, practically vacuum, the answer would be the same speed, too. Your question implies you are performing this experiment on a celestial body producing a gravity field (“straight up”).The terminal speed of an aerodynamic bomb weighing around a ton, on the atmosphere, can reach supersonic values. To obtain data on trans-sonic speeds, this was done by dropping an adequate projectile from a B-17.Terminal Velocity
A boy standing on the top of a building 10 m high throws a stone upward with a velocity of 10 m/s. What will be its velocity with which it hits the ground?
Assume velocity of upward motion is slowed at 10m/s^2. At maximum height the velocity will be zero.Time to reach max height = (V-U)/10 = (0–10)/-10 = 1 secondMax height will beh = Ut - 0.5at^2 = 10x1 - 0.5x10x1x1 = 10 - 5 = 5mtotal height H = h + 10 = 15mThe potential energy at max height = the kinetic energy at impactPE = mgh = 0.5mV^2 = KESo V = sqrt(2gH) = sqrt(2x10x15) = sqrt(300) = 17.32m/s
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 20.0 m/s?
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 20.0 m/s. The cliff is h = 42.0 m above a flat, horizontal beach (a) What are the components of the initial velocity? (b) Write the equations for the x- and y-components of the velocity of the stone with time. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.) (c) Write the equations for the position of the stone with time, using the coordinates in the figure. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.) (d) How long after being released does the stone strike the beach below the cliff? (e) With what speed and angle of impact does the stone land?
You are standing on a cliff overlooking a lake and decide to throw a tennis ball to your friends in the water below.?
You are standing on a cliff overlooking a lake and decide to throw a tennis ball to your friends in the water below. You throw the tennis ball with a velocity of 18.5 m/s at an angle of 39.5° above the horizontal. When the tennis ball leaves your hand it is 15.5 m above the water. How far does the tennis ball travel horizontally before it hits the water? In your calculations, neglect any effects of air resistance.
A person standing at the edge or a seaside cliff kicks a stone over the edge with a speed of 18m/s?
I assume the initial speed is horizontal. the time of flight is determined by how long it takes for an object, initially with no vertical velocity, to fall 52m. We find this from dist = 1/2 gt^2 or t=sqrt[2 d/g] t=sqrt[2x52m/9.8m/s/s]=3.3s the final vertical speed of the stone is gt= 9.8m/s/s x 3.3s=31.9m/s the final speed of impact is the magnitude of the velocity vector; this vector has components of 18m/s in the horizontal direction and 31.9m/s in the vertical the total final speed is sqrt[18^2+31.9^2]=36.6m/s
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of....?
h=.5gt^2 -50=.5(9.8)(t^2) a) t=3.194383 seconds v=vi+at v=0+9.8(3.194383) for just the y component the x component will remain 18 m/s the whole time v=31.304953 m/s in the y direction Pythagorean theorem 18^2+31.304953^2=c^2 c=36.110942 m/s (magnitude of resulting speed) tan^-1(31.304953/18)=60.101577 degrees b) 36.110942 m/s at 60.101577 degrees
A boy on the edge of a vertical cliff 20 m high throws a stone horizontally outwards with a speed of 20 m/s. I
Let's find the time it takes to fall 20 m d = 1/2 a t^2 t = sqrt(2d/a) = sqrt(2 x 20/10) = sqrt(4) = 2s Now let's find how far the rock can travel horizontally for 2 second when moving at a fixed speed of 20m/s. The speed is fixed because no force is exerted on the rock in the horizontal direction. d = vt = 20 m/s x 2 s = 40 m So the answer is b) 40m.
While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 7.57 m?
Sorry for the above question heres the real question While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 7.57 m/s. The stone subsequently falls to the ground, which is 14.1 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.81 m/s2.