Quadratic equations help!

Question

A ball is thrown straight up from a bridge over a river and falls into the water. The height,h, in metres, of the ball above the water t seconds after being thrown is approximately modelled by the relation h(t)=-5t²+10t+35

a)maximum height of ball above water?

b)how long does it take for the ball to reach its maximum?

c)after how many seconds does the ball hit the water?

Answer 1

this seems like a question from calculus:

 

when the ball is at its maximum height, the slope of the tangent would be zero

i.e. dh(t)/ dt = 0

Now, dh(t)/ dt = d/dt(-5t2+10t+35) = -10t + 10 = 0

=> t =1 second

meaning that after 1 sec, ball will reach it's maximum height (answer for b)

 

height at t = 1 sec would be maximum height, i.e

h (t =1 ) = -5 + 10 + 35 = 40 meters

 

(c) when ball hits the water, its height from the water would be zero, i.e. solve the following quadratic:

 

0 = -5t²+10t+35

to get the time t, when ball hits the water.

 

HTH,

 

Ved