Question

It is observed that an object that is thrown with projectile motion on a horizontal plane and reaches a maximum height of 38.5m and travels horizontally a distance equivalent to four times the maximum height. Determine the launch velocity (in m/s).

Answer 1

vertical problem:
v = Vi - g t
at top v = 0
g t = Vi
h = 0 + Vi t - (g/2) t^2
38.5 = g t^2 - (1/2) g t^2 = 4.9 t^2 on earth
t^2 = 38.5/4.9
t = 2.8 seconds upward (so 5.6 seconds in the air)
Vi = 9.8 t = 9.8 * 2.8 = 27.4 m/s
==================
Now horizontal problem x = 4 *38.5 = 154 meters
time = 5.6 seconds
so
Vx = 154/5.6 = 27.5 m/s
LOL --> launch at about 45 degrees :) v = about 1.41 * 27.45
v^2 = 27.5^2 + 27.4*2

Answer 2

yes, the 45° angle is expected. Since we know the vertex of the parabola is at (2h,h) and y=0 at 0 and 4h,
y = h - 1/(4h) (x-2h)^2
y' = -1/(2h) (x-2h)
tanθ = y'(0) = 1

Now you have another handy fact to recall, which could save you some calculation in the future.