# A golf ball is hit at angle θ to the horizontal, with initial velocity u. Stating an assumption, show that the horizontal distance travelled by the ball is directly proportional to u^2.

If we assume there is no air resistance, this means that the horizontal component of the ball's velocity won't change. Resolving horizontally we can see that the horizontal velocity = ucos(θ). We can then use this to work out how far it travels using distance = speedtime, if we know how long the ball travels for. Resolving vertically, we can see that the ball's vertical velocity is usin(θ). We can use SUVAT equations to find how long the ball takes to fall to the ground, which is also how long the ball travels horizontally for. s = 0, as the ball is starting at the ground and is going to end back up there u = usin(θ) v = ? a = -g (the negative shows that it's downwards, as we're taking 'up' to be the positive direction) t = ? We can use s = ut + 0/5at^2 here. s = 0, so this simplifies to 0 = usin(θ)t - 0.5gt^2. We can rearrange this to: usin(θ)t = 0.5gt^2, which simplifies to t = 2usin(θ)/g. We can plug this into the speed equation to give distance = 2usin(θ)/g * ucos(θ). This gives distance = 2u^2 * sin(θ)cos(θ)/g As everything except from u is a constant on the right hand side, we can therefore say that distance is directly proportional to u^2.

Answered by Douglas B. Physics tutor

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