James kicks a soccer ball into the air at a velocity of 64 feet per second.

Question

How long will it take a ball to reach its max height when kicked into the air at a velocity of 64 feet per second? What is that height? What is the range of the function of its height h in feet after t seconds? The function is h=-16t²+64t+2

Answer 1

step1 : Understand the given input data

h=-16*t^2+64*t+2

The general form of this equation is -(1/2)g.t^2+v*t+h0

g-The acceleration( gravitational).ft/sec^2. In the equation it gets negative sign because it is pulling the ball down..

v- the velocity .ft/sec.

h0- ft The initial height from the ground level.

Step 2: Determining the time ,maximum height reached.

To get the maximum speed.

determine dh/dt dh/dt=-32t+64

equate it to zero -32t+64=0

solve for t t=64/32=2 seconds.

step 3. Determining the max height.

h(2)=-16*2^2+64*2+2=-64+128+2=66 ft.

step 4 The range determination.

The ball can reach a minimum height h=0.i.e ground level.

So the range is [0,66] ft.

the time it reaches ground can be determined byequation h to zero.

-16t^2+64t+2=0

t(h=0)=(-64-sqrt(64^2+4*2*16))/(-32)= 4.0310 seconds

t(h=0)=(-64+sqrt(64^2+4*2*16))/(-32)= -0.0310 seconds

Time cannot be negative so t(h=0)=4.031 is chosen.

This is the total time of the movement.

Answer 2

The general equation, in feet, is: h = y + vt + (1/2)gt² where y is the initial heights (the ball is 2 feet off the ground), v is the initial velocity (64 ft/sec), and g = acceleration due to gravity (-32 ft/sec² downward).

The height function provided, h=-16t²+64t+2, includes the 64 feet/sec kick straight up into the air.

When the ball is kicked, it goes up, then back down. The time it takes to go up is the same as the time it takes to come back down. Since it starts at 2 feet above the ground, we solve the height equation for 2 feet. Once we have the total time for the up and down trip, we simply cut that time in half to determine the time the ball was at it's maximum height.

h=-16t²+64t+2

2=-16t²+64t+2

-16t²+64t=0

-16t(t-4)=0

Solving gives:

t=0 and 4 sec.

t=0 is the initial time when the ball was kicked.

t=4 is the round trip time

Cutting t=4 in half, t=2 is the time the ball reached it max. height.

Q: How long will it take a ball to reach its max height?

A: 2 sec.

Q: What is that height?

A: 66 Feet

Substituting t=2 sec gives:

h=-16t²+64t+2

h=-16(2)²+64(2)+2

h=-64+128+2=66 feet

Q: What is the range of the function of its height h in feet after t seconds?

A: [0,66] feet

h=-16t²+64t+2

Max h = 66 feet

Min h = 0 feet (on the ground)