Question

A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next, he gives it the same initial speed but changes the angle to
40.00˚. What effect does this have on the horizontal range? Watch significant figures! It matters in this

Answer 1

Answer:

Explanation:

Let 100 m/s  be the velocity of projection.

So horizontal component

= 100 cos42

= 74.31 m /s

Vertical component = - 100 sin 42 . in upward direction

66.91 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = - 66.91 t  + .5 x 9.8 x t²

4.9 t² -  66.91 t - 2.1 = 0

t = 13.685 s

Horizontal distance covered

= 13.685 x 74.31

= 1016.93 m

If angle of projction is 40°

So horizontal component

= 100 cos40

= 76.60 m /s

Vertical component = - 100 sin 42 . in upward direction

64.27 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = -76.60 t  + .5 x 9.8 x t²

4.9 t² -  76.60 t - 2.1 = 0

t = 15.659  s

Horizontal distance covered

= 15.659 x 76.60

= 1199.49  m

So horizontal range is increased , if angle of projection is increased .