u = 50 cos A
Vi = 50 sin A
240 = u t = 50 t cos A
so t, time in air, = 24/(5cosA)
time rising = t/2 = 12/(5 cos A)
v = Vi - g t rising
0 = 50 sin A - 9.81(12)/(5 cos A)
250 sin A cos A = 9.81(12)
now use trig to get angle A
for second part you have two values of Vi = 50 sin T
m g h = (1/2) m v^2
h = Vi^2/(2g)
A golfball with an initial speed of 50 m/s lands exactly 240 m downrange on a level course.
a) Neglecting air friction, what two projection angles would achieve this result?
b) What is the maximum altitude reached by the ball, using each of the two angles from Part A?
1 answer
1 answer