A punter can kick a football with an initial velocity of 48 feet per second.

Question

a) How many seconds will it take for the foot ball to reach maximum height

b) what is the maximum height of the football

Answer 1

Hello, Lielle

Given the details in your question, I think we've got to make some assumptions in order to solve it.

First, we aren't given any information about the ball's initial height, so we can assume that it is starting at ground level, or 0 feet.

Second, the problem says that the ball begins with an initial velocity of 48 ft/sec. Since it doesn't tell us what angle the ball is kicked at, and since this problem is about how high the ball travels, we have to assume that the ball is traveling straight up into the air.

Lastly, this problem says nothing about air resistance, so we can ignore any friction that the ball might experience as it moves through the air.

So, we know that the ball is kicked straight up at 48 ft/sec and we can ignore air resistance. If that's the case, the key to solving this problem is the force of gravity, which is the only force that is going to slow down the ball and cause it to fall back to earth. The force of gravity has an acceleration of 32 ft/sec².

So, with that out of the way, we can start to do some math. Let's just reiterate what we do know:

Ho = 0 ft Vo = 48 ft/sec A = 32 ft/sec²

Now, in order to solve this, we have to rely upon an equation of motion which relates height, velocity, and acceleration. H(t) = Ho + Vo*t + .5A*t²

Plugging in our information we get: H(t) = 48t - 16t² (Acceleration here is negative because it is working against the ball's upward movement.)

So, we've got to find two things: How long it takes the ball to get up to its maximum height, and what that height is. If this is an algebra class, I propose we solve it the following way.

Based upon the equation of motion, you know that the ball is going to travel in an arc in the shape of a downward pointing parabola. (Try plotting it for yourself on a graph where the y-axis is height and the x-axis is time.)

The vertex of that parabola occurs when the ball is at its maximum height, which will also be exactly half-way through the ball's flight. This means that it will take the ball the same amount of time to fly up to its maximum as it does to fall back down to the ground. Now, we still don't know how long it takes to get to the maximum height, but we do know that it will be exactly one half of the time that the ball is in the air - which is something we can find out.

To do so, we just need to find when H(t) = 0 (i.e. the x-intercepts of the equation).

So, H(t) = 48t - 16t²

1. To find its intercepts, we set H(t) = 0: 0 = 48t - 16t²
2. Then we isolate for the variables on the right side:
3. Both terms are divisible by 16t, so we can factor that out: 0 = 16t*(3 - t)
4. Now to find the first solution, divide both sides by (3 - t): 0/(3 - t) = 16t*(3 - t) / (3 - t)
5. This reduces to 0 = 16t
6. Again, divide both sides by 16 to get the variable by itself: 0/16 = 16t/16
7. This leaves us with 0 = t
8. But, we already new this solution. The ball is on the ground when the kick begins, t = 0.
9. So, let's get that second solution. Divide by 16t this time.
10. 0/16t = 16t*(3-t) / 16t which becomes 0 = 3-t
11. Add t to both sides to isolate the variable: 0 + t = (3-t) + t
12. t = 3 So, the ball hits the ground after 3 seconds.

Now, knowing how long the whole journey takes, we can find the time to the maximum height, which is exactly half-way, or (1/2)*t = (1/2)*(3) = 1.5 seconds.

Alright, that's the first answer to our problem. Now, to find the maximum height, all we have to do is substitute this solution (1.5 seconds) into the equation for height to know where the ball is at that time.

H(1.5) = 48*(1.5) - 16*(1.5)² which becomes H(1.5) = 72 - 36 = 36

So, it takes 1.5 seconds for the ball to reach a maximum height of 36 feet above the ground.

Hope this helps.

Daniel

Answer 2

We have a formula h(t) = -16t^2 + vt

v = 40 f/s

then h(t) = -16t² + 48t

The maximum of the parabola is the vertex.

Find the vertex of -16t^2 + 48t = 0

x = -b/2a = - (48/2*-16) = 3/2 = 1.5

So, after 1.5 second the ball reach maximum height

Find the maximum height, you plug t = 1.5 in to the equation h(t)

h(1.5) = -16(1.5)² + 48(1.5) = -16(2.25) + 72 = 36

So, the maximum height is 36 feet