Question

3. The height of a projectile object fired into the air can be modeled by h(t) = -16t² + 48t + 3, where his

height, in feet, and t is time, in seconds. What is the maximum height the projectile will reach?
A. 3 feet
B. 35 feet
C. 39 feet
D. 105 feet

Answer 1

C. 39 feet

Explanation:

`{ \tt{h(t) = - 16 {t}^(2) + 48t + 3 }}`

- Let us find the limits of the function

`{ \tt{ \frac{d \{h(t) \}}{dt} = - 32t + 48 }} \\`

- So, let us differentiate for the second time;

`{ \tt{ \frac{d {}^(2) \{h(t) \} }{dt {}^(2) } = - 32}} \\`

- So, at maximum height; d{h(t)}/dt is 0;

`{ \tt{ - 32t + 48 = 0}} \\ { \tt{ - 32t = - 48}} \\ { \tt{t = 1.5 \: seconds}}`

- Therefore; maximum height is;

`{ \tt{h(t) = - {16t}^(2) + 48t + 3}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ { \tt{h(1.5) = - 16(1.5) {}^(2) + 48(1.5) + 3}} \\ { \tt{h(1.5) = - 36 + 72 + 3}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ { \tt{h(1.5) = 39 \: feet}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:`