a cricketer can throw a ball to maximum horizontal distance of 160m .Calculate the maximum vertical height to which he can throw the ball .
g=10ms² please solve with proper steps
Answers
Answered by
14
Answer:
Maximum Range of a projectile is u^2/g, when projected 45° with the horizontal.
Given u^2/g = 160m
If the person throws the ball vertically up
He'll throw with his maximum speed 'u'
Then Maximum Height reached is u^2/2g
Which is half of the maximum Range
= 160m/2 = 80m.
The cricketer can throw the ball 80m high
Answered by
4
Answer:80m/s^-1
Explaination:
Maximum Range of a projectile is u^2/g, when projected 45° with the horizontal.
Given u^2/g = 160m
If the person throws the ball vertically up
He'll throw with his maximum speed 'u'
Then Maximum Height reached is u^2/2g
Which is half of the maximum Range
= 160m/2 = 80m.
The cricketer can throw the ball 80m high....
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