Question

A person throws a ball horizontally from the top of a building that is 26.1 m above the ground level The ball lands 112 m away from the base of the building Neglect air resistance and useg = 9.81 m/s ^ 2

Answer 1

v = 5.21 m/s

Step-by-step explanation:

Let the initial speed be v.

The top of a building is 26.1 m above the ground level and the ball lands 112 m away from the base of the building.

We need to find the initial velocity of the ball.

Let it was at rest initially. Using the second equation of motion to find time of flight.

`d=ut+(1)/(2)gt^2\\\\t=\sqrt{(2d)/(g)} \\\\t=\sqrt{(2* 26.1)/(9.81)} \\\\t=2.3\ s`

No, using the formula of speed.

`v=(d)/(t)\\\\v=(12\ m)/(2.3\ s)\\\\v=5.21\ m/s`

So, the initial velocity of the ball is 5.21 m/s.