# A-Level motion question

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#1
Question: A Ball is thrown straight up with speed "u" from a point "h" meters from the ground. Show that the time taken for the ball to strike the ground is
(v/g) [ 1 + sq. root (1 + (2hg/v^2))

So far using s = -ut + 1/2at^2.
or h = -ut + 1/2gt^2 I use this equation to solve for t. So reformat it to suit the Quadratic equation so.
1/2gt^2 - ut - h = 0 (Multiply by 2)
gt^2 - 2ut - 2h = 0

then using the quadratic equation - (note am taking upward velocity as a negative so downward is positive)

2u +- squareRoot(2u^2 - 4* g * -2h) / 2g which simplifies to
2u +- squareRoot(2u^2 + 8gh) / 2g
2u +- 2squareRoot(u^2 + 2gh) / 2g
u +- squareRoot(u^2 + 2gh) / g

so far I understand all the simplifying to this point. But am not sure on the rest. Am not sure how you're able to simplify it all the way to (v/g) [ 1 + sq. root (1 + (2hg/v^2)). Where does the "1 +" come from?

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2 years ago
#2
(Original post by zattyzatzat)
Question: A Ball is thrown straight up with speed "u" from a point "h" meters from the ground. Show that the time taken for the ball to strike the ground is
(v/g) [ 1 + sq. root (1 + (2hg/v^2))

So far using s = -ut + 1/2at^2.
or h = -ut + 1/2gt^2 I use this equation to solve for t. So reformat it to suit the Quadratic equation so.
1/2gt^2 - ut - h = 0 (Multiply by 2)
gt^2 - 2ut - 2h = 0

then using the quadratic equation - (note am taking upward velocity as a negative so downward is positive)

2u +- squareRoot(2u^2 - 4* g * -2h) / 2g which simplifies to
2u +- squareRoot(2u^2 + 8gh) / 2g
2u +- 2squareRoot(u^2 + 2gh) / 2g
u +- squareRoot(u^2 + 2gh) / g

so far I understand all the simplifying to this point. But am not sure on the rest. Am not sure how you're able to simplify it all the way to (v/g) [ 1 + sq. root (1 + (2hg/v^2)). Where does the "1 +" come from?

To start, a is needed rather than a as and subtracting it would give a negative time. Then  1
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