Can some explain how to do this problem? I tried using derivatives...but...I keep getting a different answer from what my teacher told me!
A projectile is fired directly upward with an initial velocity of 144 ft/sec and its height (in feet) above the ground after t seconds is given by s(t). Find the (a) the velocity and acceleration after t seconds (b) the maximum height and (c) the duration of the flight.
s(t)= 144-16t^2
Your equation is wrong.
s(t)=144t-16t^2
s'=velocity=144-32t
s"=acceleration=-32ft/s^2
Now try it.
To solve this problem, we first need to find the velocity and acceleration as a function of time.
(a) Velocity after t seconds:
The velocity of the projectile can be found by taking the derivative of the height function, s(t), with respect to time.
s(t) = 144 - 16t^2
Now, take the derivative of s(t):
s'(t) = d/dt (144 - 16t^2)
= 0 - 32t (the constant 144 does not have a derivative with respect to t)
So, the velocity after t seconds is given by:
v(t) = s'(t) = -32t
(b) Maximum height:
To find the maximum height, we need to determine when the projectile reaches its highest point. At this point, the velocity will be zero.
Set v(t) = 0:
-32t = 0
Solving for t, we find that t = 0. Therefore, the projectile reaches its highest point at t = 0.
Now, substitute this value of t into the height function to find the maximum height:
s(0) = 144 - 16(0)^2
= 144
So, the maximum height of the projectile is 144 feet.
(c) Duration of the flight:
To find the duration of the flight, we need to determine when the projectile hits the ground. At this point, the height will be zero.
Set s(t) = 0:
144 - 16t^2 = 0
Solve this equation for t using the quadratic formula:
t = ( -b ± √(b^2 - 4ac) ) / (2a)
For our equation, a = -16, b = 0, and c = 144. Plugging in these values, we get:
t = ( -0 ± √(0^2 - 4(-16)(144)) ) / (2(-16))
= ( ± √(0 + 9216)) / ( -32)
= ( ± √9216 ) / ( -32)
= ( ± 96 ) / ( -32)
Therefore, the duration of the flight is t = 3 seconds (taking the positive value since time cannot be negative).
So, to summarize:
(a) The velocity after t seconds is v(t) = -32t.
(b) The maximum height of the projectile is 144 feet.
(c) The duration of the flight is 3 seconds.