Vi = initial speed up
u = constant horizontal speed
find t, time in air
Vi = 135 sin 18.3 = 39.25
u = 135 cos 18.3 = 128.2
d = u T
find T the time in air
v = Vi - 9.8 t
at top where v = 0 and t=T/2
0 = 39.25 - 9.8 t
t = 4.005 s
so T = 2 t = 8.01 seconds in the air
d = u t = 128.2 * 8.01 = 1027
which is 1030 to 3 figures
A batter hits a ball, causing it to leave the bat at 135 m/s and at an angle of 18.3 degrees from the horizontal. If the baseball field is level, and ignoring the height at which the batter hits the ball, how far (horizontally, in meters) away from the batter does the ball hit the ground? Use g = 9.8 m/s2; take up and forward as positive and express your result to three significant digits.
1 answer
1 answer