Question

A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height above ground?

Answer 1

Answer:

Step-by-step explanation:

given  

A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high....  

We know that from the equation of motion,

h = u t + 1/2 at^2

So h = height, u is initial velocity and t is time, a is acceleration  

Now for the ball thrown upwards  

h = 25 t – 1/2 gt^2

25 t – h = 1/2 gt^2

Now for the ball dropped downwards

15 – h = 1/2 gt^2

Comparing both the equations we get

25t – h = 15 – h

25 t = 15

So t = 15/25

Or t = 0.6 secs

The balls will be at same height (13.2 m) when t = 0.6 secs.

(h = 25 x 0.6 - 0.5 x 10 x 0.6^2 = 15 - 1.8 = 13.2 m)

(15 - h = 1.8 or h = 13.2 m)