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chegg a stone is thrown downward with a speed of 5 m/s off a building 350 m high. find the height of the stone at time t.

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use the following suvat equation:

s = ut + 0.5at^2

where s is the displacement, u is the initial velocity, a is the constant acceleration, and t is the elapsed time.

sub given values in:

a = g (acceleration due to gravity) which has a value of 9.8 m/s^2 to 2.s.f at sea level.

s = 5t + 0.5(9.8)t^2
s = 5t + 4.9t^2

s is the displacement travelled by the stone at time t, so the height of the stone will be the initial height (350m) minus s.

Therefore the height of the stone at time t can be represented by this formula:

Height = -4.9t^2 -5t + 350 (in metres)

This also happens to be a quadratic equation, with solutions of t that are valid for values of t that are greater than or equal to zero.
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