s = initial speed
u = horizontal speed = s cos 26 = .899 s
Vi = initial speed up = s sin 26 = .438 s
It is not clear to me what your units are. If feet use g = 32 ft/s^2. If meters use g = 9.81 m/s^2. I assume from the 355 that you are using feet.
u is constant during the flight
d = u t
355 = .899 s t
so t = 395/s
now the vertical problem
for half of time t the ball is going up and the vertical speed is zero at the top
v = Vi - 32 * time rising
0 = .438 s - 32 (t/2)
16 t = .438 s
t = .0274 s
so we have two equations for t
395/s = .0274 s
s^2 = 14429
so
s = 120 ft/s
now for height
t = .0274 s
t = 3.29 seconds in the air
so 1.65 seconds rising
initial speed up = 120 sin 29 = 58.2
so average speed up = 58.2/2 = 29.1
h = 1.65 * 29.1 = 48 feet
A batted baseball leaves the bat at an angle of 26.0^\circ above the horizontal and is caught by an outfielder 355{\rm ft} from home plate at the same height from which it left the bat.
What was the initial speed of the ball?
How high does the ball rise above the point where it struck the bat?
1 answer
1 answer