A ball is thrown from the top of a building and falls to the ground. Its path is approximated by the relation h=-5t^2+5t+60.

Question

h is the height above the ground in meters and t is the elapsed time in seconds.

a) How tall is the building?

b) When will the ball hit the ground?

c) When does the ball reach its maximum height?

d) What is the maximum height?

e) Where would the ball be at 1.75 seconds? Round to the nearest meter.

f) Explain what you would do to calculate the time(s) when the ball reaches the height of 59 meters. How many solutions would there be? Why?

Answer 1

h=-5t^2 +5t +60

the building is 60 meters high

0=-5t^2 +5t +60

0 =t^2 -t -12

(t-4)(t+3) =0

t=4 seconds when it hits the ground

h' = -10t +5 =0

t=1/2 second for maximum height

h(1/2) =-5(1/2)^2 +5(1/2) +60 = -5/4+5/2+60 = 61 1/4 meters = max height

h(1.75) = h(7/4) = -5(7/4)^2 + 5(7/4) +60 = -245/16 +35/4 + 60 = -105/16+60 = 53.4375 meters

h(t) = -5t^2 +5t +60 = 59

-5t^2 +5t +1 = 0

5t^2 -5t -1 =0

t = 5/10 + or - (1/10)sqr(25+20) = 1/2 + or - (1/20)sqr45

t = 1/2 + 0.671 = about 1.171 seconds to be at 59 feet

you can ignore the negative square root, as it results in a negative value for t

the ball starts at 60 feet, goes up a little higher due to an initial velocity of 5 meters per second, and then comes down to 59 feet after a little more than 1 second.

the -5t^2 represents the effect of gravity. It really should be -4.9t^2, but apparently the problem rounded it off.