CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A ball is thrown vertically up with a certain velocity from the top of a tower of height 40 m. At 4.5 m above the top of the tower its speed is exactly half of that it will have at 4.5 m below the lop of the tower. Find the maximum height reached by the ball above the ground?

Open in App
Solution

v2v214=2g(4.5) (For top)
v21v2=2g(4.5) (For bottom)
v2(v2+2g(4.5)4)=2g(4.5)
3v24=(54)(2g)(4.5)
v2=4503=150
v=150m/s
Maximum height from tower
v2=(2g)h
h=v22g
=v220
h=7.5
Height of ball =40+7.5=47.5
1871245_1032221_ans_13260ea48b124eb78284fb9a52289f48.png

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motion Under Gravity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon