Question

a body is projected upward with velocity 100 m/sec find

(a) the maximum height reached

(b) the time taken to reach the maximum height

(c) its velocity at a height 200 m from the point of projection.

(d) the time taken to reach back the point of projection.

Take g = 10 m/sec²

Answer 1

ANSWER :-

$solution$
(a) The maximum height reached

intial upward velocity u = 100m/sec

Acceleration (a) = (-g) = -10 m/sec²

v² = u² + 2as

0 = 100² + 2(-10) H

H = 100² /2 * 10

H = 500m


(b) the time taken to reach the maximum height

t = u/g = 100/10 = 10sec

(c) velocity at a height of 200 m from the point of projection.

V² = U² + 2aS

V² = 10000 + 2(-10)200

V = ± √6000 = ± 77.46 m/sec

+ 77.46 m/sec while crossing the height upward and - 77.46 m/sec while crossing it down word

(d) the time taken to reach back the point of projection

t = 2u/g

= 2 * 100 / 10

= 20 second


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