s = 98 t - 4.9 t^2
differentiate to get velocity
v = 98 - 9.8 t
now
solve that top equation for t with s = 0 at ground
4.9 t^2 - 98 t = 0
t (4.9 t -98) = 0
obviously the height is zero when t = 0
but also when t = 98/4.9 = 20
so v = 98 - 9.8*20
= - 98 m/s
which we could have said by symmetry. It has the same speed leaving and hitting, but sign changes.
A rocket is fired into the air with an initial velocity of 98 m/s. The distance (s) the rocket is above the ground, in metres, after t seconds is given by the expression s(t) = 98t - 4.9t2. What is the rocket's instantaneous rate of change at the moment it hits the ground? Include proper units and a direction.
1 answer
1 answer