Question

A ball is thrown into the air at a velocity of 44 feet per second. Its height in feet, h after t seconds is represented by the equation h=12t^2+44t+16 after how many seconds will the ball hit the ground

Answer 1

Answer:

0.447 sec

Explanation:

velocity = distance/ time

distance 'h' is a function of time 't' given by

$h = 12t^{2} + 44t + 16$

⇒ differentiating equation

⇒ velocity = $\frac{dh}{dt} = 24t + 44$

⇒ acceleration = $\frac{d^{2}h }{dt^{2} }$ = 24 $\frac{ft}{sec^{2} }$

deceleration due to gravity = -9.8 m/$sec^{2}$ =  - 9.8 x 3.281 = - 32.154 ft/ $sec^{2}$

Net deceleration = 24 - 32.154 = - 8.15 ft/ $sec^{2}$

at maximum height v = 0

$v^{2}$ = $u^{2}$ + 2ah

⇒  0 = $44^{2}$ + 2 (-8.15)h

⇒ h = 118.72 ft⇒

118.72 = 12$t^{2}$ + 44t + 16

⇒ $12t^{2} +44t$ - 102.72 = 0

⇒  t = -44 ± $\sqrt{44^{2} - 4(12)(102.72) }$     / (2 x 12)

t = 0.447 sec