A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5 degree above the horizontal on a long flat firing range. Determine

Question

(a) The maximum height reached by the projectile,
(b) The total time in the air,
(c) The total horizontal distance covered (that is, the range), and
(d) The velocity of the projectile 1.50 s after firing.

Answer 1

do vertical problem

find Vi, initial speed up
Vi = 65.2 sin 34.5

find t where v = 0, top of curve
v = Vi - 9.81 t
0 = Vi - 9.81 t
solve for t, time at top, half the time in the air

max height
h = 0 + Vi t - 4.9 t^2

now horizontal problem
u = 65.2 * cos 34.5 the whole time
range = u*time in air = u (2 t)

(d)
horizontal component = u the whole time
vertical component = v = Vi - 9.81 (1.5)

speed = sqrt (u^2 + v^2)
tan angle to horizontal = v/u