Answer:
The maximum height is: [tex]h(t_{max})=14.39 \: u[/tex]
The ball reaches the ground in 1.79 s.
Step-by-step explanation:
We need to take the derivative and equal to zero to find the time at the maximum height.
[tex]h(t)=-16t^{2} +27t+3[/tex] (1)
[tex]\frac{dh(t)}{dt}=-32t +27=0[/tex]
[tex]t_{max} =\frac{27}{32}[/tex]
[tex]t_{max} =0.84\: u[/tex]
Now, we just need to put t(max) into equation (1) to find h(max)
[tex]h(t_{max})=-16(0.84)^{2} +27(0.84)+3[/tex]
[tex]h(t_{max})=-16(0.84)^{2} +27(0.84)+3[/tex]
[tex]h(t_{max})=14.39 \u[/tex]
If we want to get the time when the ball reaches the ground we just need to equal h(t) to zero.
[tex]0=-16t^{2} +27t+3[/tex]
Let's solve this quadratic equation.
We will get two solutions and we must choose the positive value.
t1 = 1.79 u
t2 = -0.10 u
Therefore, the ball reaches the ground in 1.79 u.
I hope it helps you!
