Question

From the top of a tower of height 100m a ball is dropped and at the same time another ball is projected with a velocity 25m/s from the ground.find when and where the two ball will meet.​

Answer 1

Answer:

▪Time when they meet➡after 4 sec

▪Distance where they meet ➡ 20m aboub the ground

Explanation:

▪Here height of tower is 100m;

▪Here acceleration is equal to acceleration due to gravity=g

FOR PART ONE➡

▪When stone is thrown upward, action is against gravity so,

a=-g

S=ut-$\frac{1}{2}$gt²

S=ut-$\frac{1}{2}$9.8t²

S=25t-5t²

FOR PART TWO➡

▪Here initial velocity of stone is zero.

▪Here g is positive because motion of stone is downward.

S'=ut+$\frac{1}{2}$gt²

S'=0t+$\frac{1}{2}$9.8t²

S'=5t²

Since total distance ➡ 100m

S+S'=100m

25t-5t²+5t²=100m

25t=100

❤t=4sec❤

Distance from top where they meet ➡

S'=5t²

S'=5(4)²

❤S'=80m❤

Distance from ground➡

100-80=20m

They will meet after 4 sec at 20m aboub the ground.