Question

If a cannonball is shot directly upward with a velocity of 160 ft per​ second, its height above the ground after t seconds is given by ​s(t)equals160 t minus 16 t squared. Find the velocity and the acceleration after t seconds. What is the maximum height the cannonball​ reaches? When does it hit the​ ground? The velocity after t seconds is ​v(t)equals 160 minus 32 t ​ft/sec. The acceleration after t seconds is ​a(t)equals negative 32 ft divided by sec squared. The cannonball reaches a maximum height of 400 ft. The cannonball hits the ground after tequals 10 sec.

Answer 1

Step-by-step explanation:

If a cannonball is shot directly upward with a velocity of 160 ft per​ second. Its height as a function of time is given by :

$h(t)=160t-16t^2$ .......(1)

t is in seconds

(a) Velocity is given by :

$v=\dfrac{dh}{dt}\\\\v=\dfrac{d(160t-16t^2)}{dt}\\\\v=(160-32t)\ ft/s$  

Acceleration is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(160-32t)}{dt}\\\\a=-32\ ft/s^2$

(b) For maximum height put $\dfrac{dh}{dt}=0$

i.e.

$160-32t=0\\\\t=5\ s$

Put t = 5 s in equation (1). So,

$h(5)=160t-16t^2\\\\h(5)=160(5)-16(5)^2\\\\h(5)=400\ ft$

(c) When the ball reaches ground, its height is equal to 0. So,

h(t) = 0

$160t-16t^2=0\\\\t(160-16t)=0\\\\t=0,t=10\ s$

Hence, this is the required solution.