Afootball is kicked at ground level with a speed of 18.0 m/s at an angle of 35.0 degrees in a horizontal direction. how much time does it take to hit the ground?

In projectile motion, the vertical velocity is independent of the horizontal velocity so we can completely ignore it. The vertical velocity is, using trigonometry, 18sin35° which is equal to 10.3 m/s.

Using a = -9.8 m/s^2 and u= 10.3 and V= 0 (because at a point, the object will be hovering in the air with 0 speed before coming back down), we have

v = u + at
=> (v-u)/a = t
=> (0-10.3)/a = t
=> -10.3/-9.8 = t
=> t = 1.05 s

the time taken to come down will be the same as the time taken to go up so total time before ball hits the ground is double the value we just found
which is 2 x 1.05 = 2.10 seconds

X- direction v = 18 cos 35 v = 14.745 m/s

y- direction initial velocity = u = 18 sin 35 = 10.324m/s

a = 9.8 m/s^2 final velocity = v = 0
t=(v−u)÷a

t = (0 - 10.324) / -9.8 = 1.05s

Since this is the time it takes to just go up, we double it to know the total time.
Hence, t = 1.05 X 2 = 2.11s

It hits the ground 2.11s later.

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