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Marat540 [252]
3 years ago
14

A ball is dropped off the roof of a tall building. If the ball reaches the ground in 8 seconds, how tall is the building, in met

ers? Round to the nearest whole number (a=9.8 m/s2)
Physics
1 answer:
Ghella [55]3 years ago
5 0

Let <em>b</em> be the height of the building, and <em>y</em> the height of the ball at time <em>t</em>, given by

<em>y</em> = <em>b</em> - 1/2 <em>gt</em>²

where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity.

It takes the ball 8 s to reach the ground, at which point <em>y</em> = 0, so that

0 = <em>b</em> - 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 313.6 m

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ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself o
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Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver

Answer:

-6461.54 N

Explanation:

From Newton's Fundamental equation,

F = m(v-u)/t.................... Equation 1

Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.

Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s

Substitute into equation 1

F = 0.75(-60-52)/0.013

F = 0.75(-112)/0.013

F = -84/0.013

F = -6461.54 N

Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.

Hence the magnitude of the average force of the wall = -6461.54 N

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