Question

# A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is : (g=10 m/s2)

A
320 m
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
300 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
360 m
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
340 m
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B 300 m Let the height of the tower is h. The motion of the ball is constant accelerating motion with acceleration g=10m/s2. Initial velocity of ball, u=20 m/s Final velocity of ball, v=80 m/s According to equation of motion for constant accelerating motion, (Here, downward direction is negative and upward direction is positive) v2−u2=2(−g)(−h) ⇒ v2−u2=2gh ⇒ (80)2−(20)2=2(10)h ⇒ 6400−400=20h ⇒ 6000=20h ⇒ h=300 m

Suggest Corrections
1
Related Videos
Applications of equations of motion
PHYSICS
Watch in App