A ball is launched into the air at an angle of 32ᵒ with an initial speed of 18 m/s. Neglecting air resistance, determine (i) how high the ball will go
(ii) how long it will be in the air, and
(iii) how far it will travel

Hiii Can somebody take their precious time to answer this question?

Vy = 18 sin 32 = 18 * .5 = 9.54    vertical speed

Vx = 18 cos 32 = 18 * .848 = 15.3 m/s horizontal speed

ty = Vy / 9.80 = .973 to get to top

t = 2 ty = 1.95 sec  (total time in air

Average speed up = 9.54 / 2 = 4.77 sec

Height ball rises = 4.77 * .973 = 4.64 m

Or H = Vy t - 1/2 g t^2

H = 9.54 * .973 - 4.9 * .973^2 = 4.64 m   as found above

D = Vx t = 15.3 *  1.95 = 29.8 m       for distance traveled

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )