Physics Question!! What did i do wrong?

The time at which the ball takes to hit the ground, if there is no air friction, only depends on the y component of position, velocity and acceleration.

Knowing this, we only need to use equation y(t) = y0 + v0y*t +.5at^2, where the initial position y0 = 0, y(t) = 0 since the ball leaves the ground and hits the ground at time t, v0y being the y component of speed (v0y = v0*sin(theta)) and a being -g.

These questions usually ignore wind resistance, take sin 33, multiply by 40 gives the vertical velocity. Then use v=vo -1/2at^2, and find t. this is time to the top. (ie vertical v=1/2at^2) when a=gravity.

Then half this vertical velocity *t will give distance it rose in the air, but that's not needed cos

Then it will fall back down due to gravity, with the same deceleration it had going up, so you can probably double the time.

ok so you have to get both the vertical component of velocity here

40 sin 33 to get the vertical

40 cos 33 to get the horizontal

then you can do v=vo+at

where v=0

vo= 40 sin 33

a=9.8

t= unknown

solve for t, and thats the time it takes to get to the max height of the football, multiply this by 2 to get the answer because that time happens twice, once to go up, once to go down

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