A ball is launched from ground level at 29 ms at an angle of

A ball is launched from ground level at 29 m/s at an angle of 38 degree above the horizontal. How far does it go before it is at ground level again? 107 m. 83 m 21 m 65 m 42 m

Solution

14. as it at ground level again.

that means vertically it is at same position as initial position.

displacement in vertical, y = 0

y = uy*t +ay * t^2 / 2

0 = (29sin38)t + (-9.8)t^2 /2

t = 2(29 sin38) / 9.8 = 3.64 sec

Now in horizontal,

Dx = vx * t = (29 cos38) (3.64) = 83.3 m .........Ans

Ans(b)