Question

eddy throws a little sand bag so that is lands on the top of a vertical post that is 4m high. The post is 1.3 m away from eddy. He releases the bag from a height of 1.5m above the ground. The initail speed of the bag is v=7.5 m/s, the angle between the velocity and the horizontal is 80 degrees. you can neglect the friction due to the air.

How long does the sand bag stay in the air?

During its course in the air, the sand bag reaches a maximum height, H. Calculate H.

What is the speed of the bag just before it lands on the top of the post. (hint: mantra speed is not velocity and velocity is not speed)?

Answer 1

Considering motion along horizontal

1.3 = u cos 80* t

ut = 1.3/ cos 80 = 7.486 ... (i)

Considering motion along vertical

4 = 1.5 + u sin 80* t - 0.5*9.8* t^2

4.9t^2 = 7.486 sin 80 - 2.5

t = 0.9972 second

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b) putting value of t in... (i)

u = 7.507 m/s

height above ground

H = 1.5 + ( u sin 80)^2/(2g)

H = 4.29 m

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Horizontal component

Vx = u cos 80 = 7.507 cos 80 = 1.3 m/s

Vertical component

Vy = 7.507 sin 80 - 9.8* 0.9972 = - 2.38 m/s

Net velocity

V^2 = Vx^2 + Vy^2

V = 2.71 m/s

The magnitude of speed is = 2.71 m/s

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