Answer:
h= -20
t = -0.104 and -2.39s
Explanation:
A juggler tosses a ball into the air. The balls height, h, and the t seconds can be represented by the equation h(t)=16t^2+40t+4. Suppose the juggler missed and the ball hit the ground. Find the maximum height of the ball and the time it took the ball to reach the ground. Round all answers to the nearest hundredth
The velocity of the body at its maximum height is zero
V= dh/dt
dh/dt = 32t + 40
0 = 32t+40
-32t = 40
t = 40/-32
t = -1.25s
Substitute t = -1.25 into the expression
h(t)=16t^2+40t+4.
h(-1.25)=16(-1.25)^2+40(-1.25)t+4.
h = 16(1.5625)-50+5
h = 25-45
h =-20
Hence the maximums height is -20m
The ball will reach the ground at h = 0
Substitute
h(t)=16t^2+40t+4.
0 = 16t^2+40t+4
Divide through by 4
0 = 4t²+10t+1
Factorize
-10±√10²-4(4)/2(4)
t = -10±√100-16/8
t = -10±√84/8
t =-10+9.17/8
t = -0.83/8 and -19.17/8
t = -0.104 and -2.39s