Height in 2d projectile motion

[stlblues]

in the book i am asked to find the maximum height of a football thrown from a quater back to a receiver. The givens are the the ball starts 1.8m off the ground, travels at 20m/s, end 1.8m off the ground, and travels 30m. I NEED HELP!

 

[Redbelly98]

Welcome to Physics Forums, St. Louis Blues.

For homework questions you are supposed to list the relevant equations and show an attempt at solving the problem, before we will help you.

 

[baba89]

In order to find the maximum height of the football in this scenario, we can use the equation for projectile motion: h = h0 + v0y*t - 1/2*g*t^2, where h is the height, h0 is the initial height, v0y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (9.8m/s^2).

We know that the initial height (h0) and the final height are both 1.8m, and the initial vertical velocity (v0y) is 0m/s since the ball is thrown horizontally. We also know that the horizontal distance (x) traveled is 30m and the initial horizontal velocity (v0x) is 20m/s.

Using the equation for horizontal distance (x = v0x*t), we can solve for t, which is the time it takes for the ball to travel 30m horizontally. t = x/v0x = 30m/20m/s = 1.5s.

Now, we can plug in our known values into the equation for height and solve for the maximum height (h):

h = h0 + v0y*t - 1/2*g*t^2
h = 1.8m + 0m/s * 1.5s - 1/2 * 9.8m/s^2 * (1.5s)^2
h = 1.8m - 1/2 * 9.8m/s^2 * 2.25s^2
h = 1.8m - 10.9125m
h = -9.1125m

Since the maximum height cannot be negative, we know that there is an error in our calculations. Upon further inspection, we can see that the initial vertical velocity (v0y) should actually be 20m/s, not 0m/s. This is because the ball is thrown at an angle, not horizontally. This changes our calculation to:

h = h0 + v0y*t - 1/2*g*t^2
h = 1.8m + 20m/s * 1.5s - 1/2 * 9.8m/s^2 * (1.5s)^2
h = 1.8m + 30m - 1