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# mechanics watch

1. i can't seem to tackle the first part
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2. (Original post by IntegralAnomaly)
i can't seem to tackle the first part
Resolve the speed, 49m/s, into horizontal and vertical components.
Using the vertucal componet, find the time taken to reach max height and then to reach thre ground again.
Us this time in your eqn of motion with the horizontal velocity.
You should then have eqns/relationships involving the angle alpha to give you that final answer.
Have you done any of that yet?
3. (Original post by Fermat)
Resolve the speed, 49m/s, into horizontal and vertical components.
Using the vertucal componet, find the time taken to reach max height and then to reach thre ground again.
Us this time in your eqn of motion with the horizontal velocity.
You should then have eqns/relationships involving the angle alpha to give you that final answer.
Have you done any of that yet?
yes,but i just can't seem to get that equation.
4. (Original post by IntegralAnomaly)
i can't seem to tackle the first part

You can express the x and y displacements of the golf ball parametrically:

-3/4/15=49sin(alpha)T-g/2T^2....(1)
98=49cos(alpha)T................ .....(2)

so from (2) T=98/49cos(alpha)

sub. into (1) and simplify to give the required expression.

(I've checked it and it works).
5. (Original post by Ralfskini)
You can express the x and y displacements of the golf ball parametrically:

-3/4/15=49sin(alpha)T-g/2T^2....(1)
98=49cos(alpha)T................ .....(2)

so from (2) T=98/49cos(alpha)

sub. into (1) and simplify to give the required expression.

(I've checked it and it works).
ah yes i was using positive 3/4/15,but why is it negative
6. (Original post by IntegralAnomaly)
ah yes i was using positive 3/4/15,but why is it negative

because you are taking its initial position as the origin and 'up' as positive displacement.
7. (Original post by Ralfskini)
because you are taking its initial position as the origin and 'up' as positive displacement.
hmmm oke thanks mate

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