Respuesta :
Answer:
Maximum height is 29 ft
Time to reach the ground is 2.60 seconds
Step-by-step explanation:
We are given equation of height as
[tex]h(t)=-16t^2+40t+4[/tex]
Maximum height:
we know that at maximum height
velocity =0
so, we will find derivative
and then we can set it to 0
and we solve for t
[tex]h'(t)=-32t+40[/tex]
now, we can set it to 0
and then we can solve for t
[tex]h'(t)=-32t+40=0[/tex]
[tex]-32t=-40[/tex]
[tex]t=\frac{5}{4}[/tex]
[tex]t=1.25[/tex]
now, we can plug t into height equation
[tex]h(\frac{5}{4})=-16(\frac{5}{4})^2+40(\frac{5}{4})+4[/tex]
[tex]h(1.25)=29[/tex]
Time to reach the ground:
we know that at ground
height=0
so, we can set h(t)=0
and then we can solve for t
[tex]h(t)=-16t^2+40t+4=0[/tex]
we can use quadratic formula
[tex]t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{-40\pm \sqrt{40^2-4\left(-16\right)4}}{2\left(-16\right)}[/tex]
[tex]t=-\frac{\sqrt{29}-5}{4},\:t=\frac{5+\sqrt{29}}{4}[/tex]
[tex]t=-0.09629,t=2.596[/tex]
Since, time can never be negative
so, we will only consider positive time
[tex]t=2.596[/tex]
Answer:
1. hmax=29m
2. 1.25secs
Step-by-step explanation:
A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40t+4 . Suppose the juggler missed and ball hit the ground . Find the maximum height of the ball and time it took to reach the ground. Round all answers to the nearest hundredth solution , at maximum height the final velocity will be zero
h(t)= -16t^2+40t+4 .................1
differentiation of h
dh/dt=0=-32t+40.................2
-40=-32t
t=1.25secs
to get the maximum height reached we juxtapose the value of t into the equation 1
h(1.25)=-16(1.25)^2+40*1.25+4
hmax=29m
2.if the ball takes 1.25secs to get to the maximum height ,it will take 1.25secs to it te round again
Therefore the time it took the ball to it the ground again is 1.25secs
