Question

A stone is projected vertically up with a velocity of 40m/s find a) the maximum height reached b) the height at the end of 2second c) velocity at the end of 3 second and d) the total time for which the stone was in the air​

Answer 1

Explanation:

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Answer 2

Answer: (a) 80m

               (b) 60m

               (c) 10 m/s

Explanation:

Initial velocity = u = 40 m/s

When the ball reaches the maximum height the velocity of ball will become zero.

∴v = 0

a = -g = -10m/s^2

Let time taken to reach be t

Maximum height = s

By $v^{2}$ - $u^{2}$ = 2as

$0^{2}$ - $40^{2}$ = 2(-10)s

-1600 = -20s

s = 80 m

To get the height at the end of 2 seconds, we will use

s = ut + 1/2at^2

Let the height at the end of 2 seconds be h

h = 40(2) + 1/2(-10)(2)^2

h = 80 - 20

h = 60m

First to check at what time the stone starts coming down we will apply formula v = u + at

0 = 40 + (-10)t

t = 4s

So at 3 seconds the stone would be going upwards

To get the velocity at the end of 3 seconds we will use

v = u + at

v = 40 + (-10)(3)

v = 40 -30

v = 10 m/s  

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