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Vikki [24]
3 years ago
14

You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

a = g = 9.8 m/s^2

Part b)

Now from kinematics equation we will have

y = v_i t + \frac{1}{2} at^2

now we have

y = 25 m

so from above equation

25 = 0 + \frac{1}{2}(9.8 )t^2

t = 2.26 s

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

v_x = vcos\theta

v_x  = 20 cos65 = 8.45 m/s

v_y = vsin\theta

v_y = 20 sin65 = 18.13 m/s

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

v_f - v_y = at

0 - 18.13 = (-9.8) t

t = 1.85 s

so it will take 1.85 s to reach the top

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