Whoa never done physics looks hard tho
Answer:
vx = 65 yd/3 sec = 21.7 yd/sec since horizontal speed is constant
vy = g t = (32 / 3) yd/sec^2 * 1.5 sec = 16 yd/sec where 32/3 is the acceleration due to gravity in yds / sec^2 and 1.5 is the time to travel each way in the vertical direction
V = (vx^2 + vy^2)^1/2 = (21.7^2 + 16^2)^1/2 = 27 yd/sec
tan theta = vy/vx = 16 / 21.7 = .737 theta = 36.4 deg
You can check using the range formula:
R = v^2 sin (2 theta) / g = 27^2 * .955 / (32 / 3) = 65.3 yds
The difference from 65 yds may be rounding error.
Mass and velocity of course
Answer:It depends on the initial velocity of the projectile and the angle of projection. The maximum height of the projectile is when the projectile reaches zero vertical velocity. ... The horizontal displacement of the projectile is called the range of the projectile and depends on the initial velocity of the object
