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A body is thrown vertically upward with velocity of 90m/s.Calculate:-

a)Maximum height reached.

b)Time taken to reach the highest point.

c)Velocity at a height of 196m from the point of projection.

d)Velocity with which it returns to ground.

e)Time taken to reach the ground

Initial velocity of the body in the upward direction, u = 90 m/s

(a) Final velocity of the body at the highest point is zero.

Maximum height attained by the body,

${v}^{2}-{u}^{2}=2as\phantom{\rule{0ex}{0ex}}\Rightarrow 0-{\left(90\right)}^{2}=2\times \left(-10\right)\times h\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{-8100}{-20}=405\mathrm{m}$

(b) Time taken to reach the highest,

$v=u+at\phantom{\rule{0ex}{0ex}}\Rightarrow 0=90-10\times t\phantom{\rule{0ex}{0ex}}\Rightarrow t=10\mathrm{s}$

(c) Let

*v*' be the velocity of the body at height 196 m.

So,

$v{\text{'}}^{2}-{u}^{2}=2as\phantom{\rule{0ex}{0ex}}\Rightarrow v{\text{'}}^{2}-{\left(90\right)}^{2}=2\times \left(-10\right)\times 196\phantom{\rule{0ex}{0ex}}\Rightarrow v{\text{'}}^{2}=-3920+8100=4180\phantom{\rule{0ex}{0ex}}\Rightarrow v\text{'}=\sqrt{4180}=64.65\mathrm{m}/\mathrm{s}$

(d) It will return to the ground with the same speed with which it was projected upwards but the its direction is opposite to the initial direction.

Thus, the body reach the ground with a velocity of -90 m/s.

(e) Time taken to reach the ground = 2 x (Time taken to reach the â€‹highest point)

Time taken to reach the ground = 2 x 10 = 20 s.

Regards,

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