Question

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18s. What is the value of v? (take g = 10 m/s2)(a) 75 m/s (b) 55 m/s(c) 40 m/s (d) 60 m/s

Answer 1

the answer is in the image. hope it helps

Answer 2

Answer:

A)  75 m/sec

Explanation:

t = 0 (Given)

Time when ball is dropped = 6 sec (Given)

Time when balls meet = 18 sec (Given)

The vertical distance traveled by ball one by t = 18 sec

= s1 = u × t + 1/2 gt²

= s1 = 0 + 1/2 g × 18²

= 162 g

Distance traveled by ball two in 12 secs. from t=6 sec to 18 sec. 

= s2 = v × 12 + 1/2 g × 12²

= 12v+72 g

s1 = s2

= 12 v = 90 g

= v = 7.5 g

or  75 m/sec

Thus, when the two balls meet at t = 18s the value of v is  75 m/sec