chain rule - calculus

Question

The height of an object launched straight up is

h(t) = 100 + 64t − 16t²

where t is measured in seconds and h is measured in feet. Answer the following questions.

All answers must have correct units and must be accurate to two decimal places.

How high is the object when t = 5

seconds?

__________

How fast is it moving when t = 5

seconds?

_____________

When is the height exactly 5 feet? (Assume t > 0.)

______________

When is the velocity exactly 5 ft/sec?

____________

Answer 1

h(t) = 100 + 64t - 16t^2

h(5) = 100+64(5) +-16(5)^2 = 100+ 320 - 400 = 20 feet high

velocity = h'(t) = 64-32t

v(5) = h'(5) = 64-32(5) = 64- 160 = -96 feet per second velocity in 5 seconds

5 = 100 + 64t - 6t^2

6t^2 -64t -95 = 0

t= 64/12 + or - 1/12)(sqr(64^2 +24(95)) = (64+or - 79.81228)/12 = 11.984 or -1.3177 seconds

It reaches 5 feet after 11.98 seconds

v = 5 = 64 -32t

32t = 64-5 = 59

t = 59/32 = 1.84 seconds to reach velocity = 5 feet per second