a) and c)
It keeps going down from 39.2 meters
b) when is h = 0?
4.9 d^2 +9.8 d - 39.2 = 0
d = -4 or 2
d is not negative, that was before you shot it
so d = 2
now really d is u t where u is the constant horizontal speed
whoever wrote this was not a physicist
it should probably be h(t) = –4.9 t^2 - 9.8 t + 39.2
and the time t comes out about 2 seconds in the air, not just two meters flight.
The path of an arrow can be modelled by the function h(d) = –4.9d 2 -9.8d+39.2,
where h(d) is the height of the arrow and d is the horizontal distance travelled,
both in metres.
a) What is the maximum height of the arrow?
b) When does the arrow hit the ground?
c) What is the initial height of the arrow?
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