Question

10.
A student is able to throw a ball vertically to
maximum height of 40 m. The maximum distance
to which the student can throw the ball in the
horizontal direction :-
(1) 40 (2)1/2 m
(2) 20(2)1/2m
(3) 20 m
(4) 80 m​

Answer 1

Answer:

Explanation:

Time of ascent=time of descent

v^2=u^2-2*40*10  ( as 'g' is negative)

as v =0(as its the max. height and at max. height final velocity is 0)

u^2=800

u=20 * root 2

Therefore the student can throw the ball in any direction but the velocity will be 20*1.4, which is 28 (approx). So, he can throw the ball this far.

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Answer 2

(4) 80 m

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