A ball is thrown into the air with an initial upward velocity of 60 ft/s. Use the equation h(t)=-16t^2+60t to answer.

Question

1. What's the max height the ball can reach?
2. After how many seconds will the ball reach the ground and max height?
3. Whats the height of the ball after 3 seconds

Answer 1

What's the max height the ball can reach? To find the max height, we should find the vertex. For an equation in standard form, the x coordinate of the vertex is t = -b/(2a) For this problem, t = -60/[(2)(-16)] t=1.875.

Now use this time to find the max height. h = (-16)(1.875)²+60(1.875) = 56.25 ft

After how many seconds will the ball reach the ground and max height?

To find when it hits the ground, set the equation equal to zero and solve.

0= -16t²+60t factor out -4t

0 = -4t(4t-15) Set the factors equal to zero and solve

4t-15=0 Add 5 to both sides

4t=15 Divide both sides by 4

t=3.75 sec

The time to get to the max height was found in the first part, it is the x coordinate of the vertex.

t=1.875 seconds

What's the height of the ball after 3 seconds Find h(3)

h = (-16)(3)²+60(3)

h = 35 ft

Answer 2

h= -16t^2+60t

h' = -32t +60 =0

t= 60/32 = 15/8 = 1.875 seconds to reach max height

max height = -16(15/8)^2 +60(15/8) = -225/4+450/4 =225/4=56.25 feet

it hits the ground when -16t^2+60t = 0

when 4t^2-15t= 0

when t(4t-15) = 0

t=0 or 15/4=3.75 seconds

in 3 seconds h(3) = -16(3^2 +60(3) = -144+180 =36 feet from the ground

another way to solve the problem is rewrite the equation in vertex form, by completing the square

h(t) = -16t^2 +60t = -16(t^2 - 60t/16) = -16(t^2-15t/4) = -16(t^2-15t/4+225/64) +225/4

=-16(t-15/8)^2+56.25

vertex= (15/8, 56.25) = maximum with 56.25 feet max height in 15/8 seconds