Finding total velocity before hitting the ground and the angle it makes with the horizontal?

Question

A golf ball is hit off the top of a hill with velocity of 35m/s at an angle of 60* with the horizontal. The hill is 7m high. I found the total time that the golf balls in the air is 5.95 seconds but i cant find out how to find the total velocity before the golf ball hits the ground and the angle it makes with the horizontal? Help?

Answer 1

Firstly I checked your value for the total time of flight for the golf ball and it is correct.

I am unsure what you mean by "total velocity" so I presumed that you mean impact velocity of the golf ball as it hits the ground).

To find out the impact velocity of the ball, we need to find the vertical and horizontal speeds of the golf ball as it impacts the ground. Then we use pythagoras' theorem to find the velocity of the ball on impact.

The horizontal speed of the golf ball is constant if air resistance is ignored (which it is). So we find the initial horizontal speed of the ball.

It is hit off the hill at 35 ms-1 @ 60 degrees. So we use the Cos ratio.

35*Cos 60 = 17.5 ms-1 (Intially it is hit with a horizontal velocity of 17.5 ms-1)

Remember, our objective is to find the vertical and horizontal speeds of the golf ball as it impacts the ground. So we can use pythagoras' theorem to find the velocity of the ball on impact.

Since the horizontal velocity of the golf ball is constant, it is travelling at 17.5 ms-1 horizontally at the start and at the end of its flight. Therefore the horizontal impact velocity of the ball is 17.5 ms-1.

Now we need to find the vertical impact velocity of the ball. It is hit off the hill at 35 ms-1 @ 60 degrees. So we use the Sin ratio.

35*Sin 60 = 30.31 ms-1 (Intially it is hit with a vertical velocity of 30.31 ms-1).

So to find the vertical impact velocity of the ball we use this formula.

v = u + at (v is final velocity. u is initial velocity, a is acceleration, t is time of flight.)

v = 30.31 + (-9.8 x 5.95)

  = -28 ms-1 (The ball hits the ground with a vertical velocity of 28 ms-1).

Now we know the vertical (28 ms-1) and horizontal (17.5 ms-1) impact velocities of the ball (draw a triangle, label the horizontal side 17.5 ms-1 and label the vertical side 28 ms-1. We now are finding the hypotenuse of the triangle, the actual impact velocity of the golf ball).

Using pythagoras' theorem

c^2 = a^2 + b^2

c^2 = 28^2 + 17.5^2

c^2 = 1090.25

c = 33.02 ms-1

This is the impact velocity of the golf ball.

To find the angle it makes with the horizontal.

tan (theta) = opposite/adjacent

tan (theta) = 28/17.5

theta = tan-1 (28/17.5)

theta = 58 degrees.

I believe that is your question answered. Sorry about it being so long but I felt like it would be a good idea to explain things properly.