posted by Jennifer on .
A ball player hits a home run, and the baseball just clears a wall 17.0 m high located 124.0 m from home plate. The ball is hit an an angle of 35° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.0 m above the ground.
(a) What is the initial speed?
(b) How much time does it take for the ball to reach the wall?
(c) Find the components of the velocity and the speed of the ball when it reaches the wall
vy,f = m/s
vx,f = m/s
vf = m/s
a. Dh=Vo^2*sin(2A) / g = 124m = Hor. dist.
Vo^2^sin(79) / 9.8 = 124,
0.9397Vo^2 / 9.8 = 124,
0.9397Vo^2 = 1215.2,
Vo^2 = 1293.2,
Vo = 36m/s @ 35 = Initial velocity.
Xo = 36cos35 = 29.49m/s.
Yo = ver. = 36sin35 = 20.65m/s
c. hmax = (Yf^2 - Yo^2) = 2g,
hmax = (0 - (20.65)^2) / -19.6 = 21.8m.
Yf^2 = Yo^2 + 2g*d,
Yf^2= 0 + 19.6*(21.8-17)=94.1,
Yf = 9.7m/s. = Vy,f.
Xo = 29.49m/s = Vx,f.
Vf = sqrt((29.49)^2 + (9.7)^2) = 31m/s.
b. t(up) = (Yf - Yo) / g,
t(up) = (0 - 20.65) / -9.8 = 2.1071s.
t(dn) = = (Yf - Yo) / g,
t(dn) = (9.7 0) / 9.8 = 0.9898s.
T = t(up) + t(dn)
T = 2.1071 + 0.9898 = 3.1s to reach te