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

The velocity and acceleration of a cannonball shot directly upwards can be calculated using known kinematic equations. The maximum height is achieved when velocity equals zero, and the time it hits the ground can be found by setting the height equation to zero.

Step-by-step explanation:

When a cannonball is shot directly upwards, the position, velocity, and acceleration at any point in time can be determined using kinematic equations. The height s(t) after t seconds is given by s(t) = 160t - 16t2. The velocity v(t) after t seconds is v(t) = 160 - 32t, and the acceleration a(t), which is due to gravity, is a constant at -32 ft/s2. To find the maximum height reached by the cannonball, we set the velocity v(t) to zero and solve for t, then substitute this value back into s(t).

To determine when the cannonball hits the ground, we solve s(t) = 0 for t. It is known that the maximum height the cannonball reaches is 400 ft and it hits the ground after 10 seconds as per the provided information.

Answer 2

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=(dh)/(dt)\\\\v=(d(160t-16t^2))/(dt)\\\\v=(160-32t)\ ft/s`

Acceleration is given by :

`a=(dv)/(dt)\\\\a=(d(160-32t))/(dt)\\\\a=-32\ ft/s^2`

(b) For maximum height put
`(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.