Question

a body is thrown vertically upward remains in air for 2 seconds . another body is thrown upward with double velocity. how lo g will it stay in air ​

Answer 1

answer : 4 seconds....

solution:

(i) for first body,

total time to return to the ground = 2 s

time taken to reach topmost point = 2/2 = 1 s

at topmost point , v = 0 m/s

a = -10 m/s^2

let initial velocity= u m/s

so,

v = u + at

=> 0 = u - 10×1

=> u = 10

(ii) for second body,

u = 2×10 = 20 m/s

a = -10 m/s^2

at topmost point, v = 0 m/s

again,

v = u + at

=> 0 = 20 - 10t

=> t = 20/10 = 2

time taken to reach the topmost point = 2 s

so, total time to return to the ground = 2 × 2 = 4 s

.........answer...

hope , it helped you.

Answer 2

Explanation:

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