wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A ball is thrown upward with an initial velocity of 100 ms1. Find the height of the ball after 15 s. Take g=10 ms2

A
375 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
425 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
395 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
275 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 375 m
Here, u=100 ms1, a=g=10 ms2
At highest point, v=0
As v=u+gt0=10010×t
Time taken to reach highest point,
t=10010=10 s
The ball will return to the ground at t=20 s
Corresponding velocity - time graph of the ball is shown in
Height attained after 15 s=Area of ΔAOB+Area of ΔBCD
=500+12(1510)×(50)=500125=375 m

flag
Suggest Corrections
thumbs-up
12
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motion Under Constant Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon