logo
or
search
upload

A ball is thrown upward with an ini tial velocity of 100 "ms"^(-1). After how much time will it return ? Draw velocity - time graph for the ball and find from the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15 s. Take g=10 ms^(-2) .

Apne doubts clear karein ab Whatsapp par bhi. Try it now.
CLICK HERE

Loading DoubtNut Solution for you

Watch 1000+ concepts & tricky questions explained!

200+ views | 100+ people like this

   
Share
Share

Answer Text

Solution :
<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/AKS_CAO_PHY_V01_P1_C03_SLV_042_S01.png" width="80%"> <br> Here , `u=100 ms^(-1), g=-10 ms^(-2)` <br> At highest point , v=0 <br> As v=u+gt `therefore` 0=100-10 x t <br> `therefore` Time taken to reach highest point , `t=100/10`=10s <br> The ball will return to the ground at t = 20 s <br> Corresponding velocity- time graph of the ball is shown in <br> (i)Maximum height attained the ball=Area of `triangleAOB` <br> `=1/2xx10xx100` =500 m <br> (ii)Height attained after 15s=Area of `triangleAOB`+Area of `triangleBCD` <br> `=500+1/2(15-10)xx(-50)=500-125`=375 m

Related Video

4:54

92.2 K+ Views | 19.0 K+ Likes

2.6 K+ Views | 100+ Likes

8:22

61.6 K+ Views | 5.9 K+ Likes

5:42

15.3 K+ Views | 900+ Likes

3:56

248.2 K+ Views | 153.9 K+ Likes

2:16

146.8 K+ Views | 130.7 K+ Likes

2:52

125.5 K+ Views | 6.2 K+ Likes

bottom open in app