When an object is thrown? upward,

Question

When an object is thrown? upward, its height? (S), in? meters, is given? (approximately) by the quadratic equation Sequals=negative 5 t squared plus vt plus h−5t2+vt+h?, where v is the upward initial velocity in? meters/second, t is the time of flight in? seconds, and h is the height above level ground from which the object is thrown. Johnny is standing on a platform 1515 meters high and throws a ball straight up as high as he can at a velocity of 2222 ?meters/second. At what time t will the ball hit the? ground? How far from the ground is the ball after 44 seconds have elapsed from the time of the? throw? (Assume that the ball is 1515 meters from the ground when it leaves? Johnny's hand.)

Answer 1

s(t) = -5t² + vt + h

 

I think there are typos in the problem statement.  I assume you meant

 

h = 15 m

v = 22 m/s

 

Then

 

s(t) = -5t² + 22t + 15

 

It hits the ground when s(t) = 0

 

-5t² + 22t + 15 = 0

 

t = [1/(2*(-5)] {- 22 ± √[22² - 4(-5)(15)]}

 

t = (-1/10) (-22 ± √784) = (-1/10) (-22 ± 28)

 

Note that in order to get a positive time, we need to use the negative square root.

 

t = (-1/10) (-22 - 28) = 5s

 

It hits the ground after 5 seconds.

 

Is 44 seconds a typo?  I will assume you meant 4s.

 

s(4) =  -5(4²) + 22(4) + 15 = 23

 

At 4s it is 23 m above the ground.