An astronaut on the moon throws a ball vertically upwards. The height, s metres, of the ball, after t seconds, is given by the equation s= 40t + 0.5at^2, where a is a constant. If the ball reaches its maximum height when t = 25, find the value of a. a=-1.6 differenciate wrt t ds/dt = 40+ at to maximise s, make the ds/dt zero.. so, a=-40/t=-40/25=-1.6

Question

An astronaut on the moon throws a ball vertically upwards. The height, s metres, of the ball, after t seconds, is given by the equation s= 40t + 0.5at^2, where a is a constant. If the ball reaches its maximum height when t = 25, find the value of a.

<explain the process below.>

a=-1.6
differenciate wrt t
ds/dt = 40+ at
to maximise s, make the ds/dt zero..
so, a=-40/t=-40/25=-1.6

anonymous · Answer

At its maximum height, the ball will have a velocity of zero. Thus, (since velocity = ds/dt ) we differentiate with respect to t (as this is the independent variable that is changing). So, $$\frac{ds}{dt} = \frac{d(40t)}{dt} + \frac{d(0.5at^2)}{dt}$$. The derivative of 40t is just 40. The derivative of 0.5at^2 is 0.5at/2=at (by the power rule). Added together we get ds/dt = 40 + at. Since we know that at t=25, ds/dt = 0, we set ds/dt = 0 and solve for a. 
ds/dt=40+at 
0 = 40+a*25 
-40 = a*25  
-40/25= a = -1.6