(b) How many seconds, after the ball is released, will it hit the ground?

Question

David drops a soccer ball off a building. The building is 75 meters tall.

What is the height of the ball to the nearest tenth of a meter exactly 1 seconds after he drops the ball?

Answer 1

Measure the height of the ball as positive from the ground.

At time=0, the height of the ball is 75 m.

Consider the motion of the ball from when it is dropped.

The initial vertical velocity is zero, say u=0 m/s.

The acceleration due to gravity is g = 9.8 m/s².

To determine the vertical distance traveled downward in 1 second, use the formula

s = ut + (1/2)*a*t²

That is,

s = (0 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)² = 4.9 m

Therefore the vertical height of the ball is:

Initial height - Distance descended = 75 - 4.9 = 70.1 m

To the nearest tenth of a meter, the height of the ball after 1 second is 70.1 m.

I hope the solution is clear.

Answer 2

Refer to newtons kinematic equations.

For for freefall we have d = (1/2) gt^2

g= 9.81 m/s^2

t =1s

d = (1/2)(9.81)(1)^2 =4.9m

now this is the height the ball has fallen thus far

75m - 4.9m = 70.1m, the height of the ball after 1 second