The height of a projectile launched into the air is given by h (t)= -16t^2 + 48t + 60, where h is the height of the projectile in feet above the ground and t is the number of seconds after its launch. Find the *instantaneous rate of change* in the height of the projectile at 1.5 seconds. It is supposed to be 24 feet per second. Isn't it just -32t + 48 and then putting the 1.5 in for that t?
h(t)=-16t^2 + 48t + 60
dh/dt=-32t+48, at t=1.5, then
h'=0
you are right.
yes, but dh/dt=0 at t=1.5
Don't know how they get 24. dh/dt=24 at t=3/4
To find the instantaneous rate of change in the height of the projectile at 1.5 seconds, we need to find the derivative of the height function h(t) with respect to t, and then evaluate it at t = 1.5 seconds.
Let's calculate the derivative of h(t) using the power rule of differentiation:
h'(t) = -32t + 48
Now, to find the instantaneous rate of change at t = 1.5 seconds, substitute t = 1.5 into h'(t):
h'(1.5) = -32 * 1.5 + 48
= -48 + 48
= 0 feet per second
The result obtained is 0 feet per second, not 24 feet per second as expected.
It seems that the derivative of the height function you provided is incorrect. The correct derivative of h(t) = -16t^2 + 48t + 60 is h'(t) = -32t + 48.
Therefore, the instantaneous rate of change in the height of the projectile at 1.5 seconds should be calculated as follows:
h'(1.5) = -32 * 1.5 + 48
= -48 + 48
= 0 feet per second
So, the final answer is 0 feet per second, not 24 feet per second.
To find the instantaneous rate of change of the height of the projectile at a specific time, you need to find the derivative of the height function with respect to time. The derivative represents the rate at which the function is changing at any given moment.
In this case, you correctly identified that the height function is h(t) = -16t^2 + 48t + 60. To find its derivative, apply the power rule for differentiation:
h'(t) = d/dt(-16t^2 + 48t + 60)
= -32t + 48
The derivative of the height function is h'(t) = -32t + 48.
To find the instantaneous rate of change at t = 1.5 seconds, substitute t = 1.5 into h'(t):
h'(1.5) = -32(1.5) + 48
= -48 + 48
= 0
It seems you made a mistake when substituting t = 1.5 into h'(t) as -32(1.5) + 48 would equal 0, not 24.
Therefore, the instantaneous rate of change of the height of the projectile at 1.5 seconds is 0 feet per second, not 24 feet per second.