Question

A person can throw a stone to a maximum height of h meter.The maximum distance to which he can throw the stone is​

Answer 1

Answer:

ATQ

s= h

v=0

a = -g

u = u

v²-u²=2as

u² = 0 + 2×g×h

u² = 20h

Now let the person throws the stone at an angle ∅ with ground then

Distance

R = 2u²sin∅cos∅/g

= 40h sin45°cos45° / 10 (Range is max. at 45° angle )

= 40h×1/2/10 = 2h

Maximum distance to which person can throw is 2h

Answer 2

Answer:Range Max=Zero

Explanation:becoz for maximum height angle will be 90°.Hmax= u^2 sin^2 (90)÷2g gives maximum height but Range max = u^2 sin2 (pie÷2) ÷g

Which gives Zero distance .

So when maximum is attained by the particle then Range max will be ZERO