Answer:
10.5618feet, 0.8125sec
Step-by-step explanation:
Given the height reached by the catapult expressed by the equation H=-16T^2+26T
At the maximum height, the velocity of the object is zero, hence;
v = dh/dt = 0
-32t + 26 = 0
-32t = -26
t = 26/32
t = 0.8125secs
Hence it will attain the maximum height after 0.8125second
Substitute t = 0.8125s into the expression to get the maximum height
H=-16t^2+26t
H = -16(0.8125)^2 + 26(0.8125)
H = -16(0.6602) + 21.125
H = -10.5632+21.125
H = 10.5618
hence the maximum height reached by the ball is 10.5618feet
