g = -9.8 m/s^2
v = +20 - 9.8 t
y = 10 + 20 t -4.9 t^2
a) when is y = 10 again?
10 = 10 + 20 t -4.9 t^2
t(4.9 t-20) = 0
t = 20/4.9 = 4.08 seconds
b) when is y = 0 ?
4.9 t^2 - 20 t - 10 = 0
t = [20 +/- sqrt (400+196)] / 9.8
forget negative time, that was before we started, use + sign
t = 4.53 seconds
c) velocity is 0 at the op when
0 = 20 -9.8 t
t = 2.04 seconds
d) y = 10 + 20 (1) -4.9 (1)^2
= 25.1 meters
A projectile is launched from a platform 10 meters above ground with an initial upward velocity of 20 meters per second.
Find the time when the projectile has returned to the initial height launch, find the time the projectile hits the ground, find the time when the projectile hits the ground when its velocity is 0, find the height of the projectile above ground after 1 second
1 answer
2 answers