The motion described here is a projectile motion which is characterized by an arc-shaped direction of motion. There are already derived equations for this type of motions as listed:
Hmax = vâ²sin²θ/2g
t = 2vâsinθ/g
y = xtanθ + gx²/(2vâ²cos²θ)
where
Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
vâ= initial velocity
t = time of flight
x = horizontal range
y = vertical height
Part A.Â
Hmax = vâ²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m
Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:
y = vât + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds
a = (vâ-vâ)/t
9.81 = (vâ - 30)/0.839
vâ = 38.23 m/s
Part C.Â
y = xtanθ + gx²/(2vâ²cos²θ)
Hmax + 15 = xtanθ + gx²/(2vâ²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m
