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Question from Jessi, a student:

Ben is out at the practice range hitting golf balls. How much further will a golf ball with an initial speed of 75.0 m/s go when projected at 45.0 degree than when projected at 30.0 degree?

Hi Jessi.

This is a question of projectile motion.

There are a couple of very important equations of projectile motion:

y = y0 + v0 t sinθ - ½ g t2.
x = x0 + v0 t cosθ.

Where y is the vertical location, x is the horizontal location, x0, y0 are the initial location, v0 is the initial velocity, t is the time since launch, θ is the angle between the initial trajectory and the ground and g is the constant acceleration due to gravity: 9.8 m/s2.

You can use the first equation for each value of θ to find t for each situation:
0 = 0 + 75 t sinθ - ½ g t2.

Once you have t for each angle, you can use it in the second equation, which describes the horizontal movement, to find where the ball strikes the ground.

I think you are meant to ignore special aerodynamic properties of golf balls, air resistance and how far it bounces!

Cheers,
Stephen La Rocque.

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