A ball is thrown from a building 35m tall and lands on the ground 5 m away from the building.?

Question

What was the initial horizontal velocity?

Answer 1

Assuming it was thrown horizontally....

The time it takes to fall due to the acceleration of gravity:

x = 0.5 * a * t^2

35 = 0.5 * 9.8 * t^2

7.142 = t^2

2.67 seconds = t

If it made it 5 m in 2.67 seconds, then it was travelling (horizontally) at a rate of 5/2.67 = 1.87 m/s

.

Answer 2

Horizontal velocity is vcos0 (where 0 is theta).

Where 0 is the angle at which the ball is thrown.

Max Height is 35 meters.

Ball falls at 10 m/s/s.

delta x = vot + 1/2 at^2

35 = vot + 1/2 at^2

35 = 1/2(10)t^2

35 = 5t^2

7 = t^2

2.67 = t, time until it hits the ground.

distance horizontal = v (horizontal) * t

5 = vcos0 * time

5 = vcos0 *2.67

5/2.67 = vcos0

1.87 = vcos0

Answer 3

It may have initial vertical velocity