A boy throws a ball upwards with a velocity of $9.8 m s^{- 1}$ . How high does it go?

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Solution

Given,
Initial velocity of the ball = 9.8 m/s.
Using the third equation of motion we get
$v^{2} = u^{2} + 2 a s$

$0 = 10^{2} + 2 (- 9.8) s$

$19.6 s = 100$

$∴ s = 5.10 m$

Therefore, the ball goes upto a height of $5.1 m$

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A boy throws a ball upwards with a velocity of 9.8ms−1. How high does it go?

Given,Initial velocity of the ball = 9.8 m/s.Using the third equation of motion we getv2=u2+2as0=102+2(−9.8)s19.6s=100∴s=5.10mTherefore, the ball goes upto a height of 5.1m

A boy throws a ball upwards with a velocity of $9.8 m s^{- 1}$ . How high does it go?

Given,
Initial velocity of the ball = 9.8 m/s.
Using the third equation of motion we get
$v^{2} = u^{2} + 2 a s$

$0 = 10^{2} + 2 (- 9.8) s$

$19.6 s = 100$

$∴ s = 5.10 m$

Therefore, the ball goes upto a height of $5.1 m$