Physics Question?

Question

Hello,
I have this problem that i need to solve in physics however evertyhing that I come up with doesn't add up it makes no sense for me. Here is the problem:

A placekicker must kick a football from a point 36.0m from the goal, and the ball must clear the crossbar, which is 3.05m. When kicked, the ball leaves the ground with a speed of 20m/s at an angle of 53 degrees to the horizontal. (A) By how much does the ball clear or fall short of clearing the crossbar? (B) Does the ball approach the crossbar while still rising of while falling?

So here is what I did. I asumed that the initial V was 20m/s and with it I found the time. Then i found the distance of Y using the 3.05sin53 then I found the velocity of why with the answer and with it I applied it to the equation delta y= Viyt + 1/2ayt

However since b asks if it is falling or raising I assumed it would be consiredered as a projectile. So i used a = -g. But it doesn't add up. I get -0.02m.

Was there something I did wrong?

Answer 1

Horizontal velocity is 20 cos 53° = 12.036 m/s. Round this to 12 m/s. Time required to traverse 36 m at this speed, t = 36/12 = 3 s.

Now find ball height: h = 20 sin 53° × 3 − 9.8/2 × 3² = 3.8 m. This means the ball clears the crossbar by 0.75 m, approx.

To find wether the ball is still rising or falling, find the time at which the ball reaches maximum height. At this point, vertical velocity is zero (if it weren't so, this wouldn't be the point of maximum height, neither). By definition, v − v₀ = at, and you know that vertical velocity is zero at this time; a = g, of course, and don't forget we're dealing with VERTICAL velocity.

The equation becomes

-v₀sin 53° = -gt,

whence t = v₀sin 53° / g = 20 sin 53° / 9.8 = 1.63 s.

Obviously, then, at 3 s the ball is falling.