Problema Solution
The formula d=vt-5t^2 gives the approximate distance in meters of an object above its starting point t seconds after it was thrown with an initial upward velocity of v meters per second. How long would it take a ball thrown up at an initial velocity of 55 meters per second to reach a height of 50 meters?
Answer provided by our tutors
d=vt-5t^2
v = 55 m/s
d = 50 m
t = ?
55t - 5t^2 = 50
by solving we find
t1 = 1 s
t2 = 10 s
the ball will reach height after 1 second on the way up (first time) and after 10 seconds on the way down (second time).