Guest

A ball is thrown upward and caught by the thrower after 8 seconds.How below the highest point was it 5 seconds from the start? 1m 4.9m 9.8m 19.6m PLS ANSWER FAST

A ball is thrown upward and caught by the thrower after 8 seconds.How below the highest point was it 5 seconds from the start?
  • 1m
  • 4.9m
  • 9.8m
  • 19.6m
PLS ANSWER FAST

Grade:12th Pass

1 Answers

Arun
25750 Points
6 years ago

8 second here indicates the total time taken by the ball to reach highest point and come back to the thrower. So the highest point is reached at half the time i.e. 4 seconds. take g=9.8m/s^2

v=u-gt (since ball goes upward)

at highest point v=0, time taken=4

u=v+gt

u=39.2m/s…….

v^2=u^2-2gh (ball goes upward)

at highest point v=0

h=u^2/2g

h=78.4m……..(highest point)

h=ut- .5gt^2

t=5 seconds, u=39.2

h=196 - 0.5 *9.8 *25

h = 73.5

so your answer would be= 78.4–73.5= 4.9m

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free