posted by Anonymous on .
A ball is thrown from a point 1.1 m above the ground. The initial velocity is 20.3 m/s at an angle of 32.0° above the horizontal.
(a) Find the maximum height of the ball above the ground???
(b) Calculate the speed of the ball at the highest point in the trajectory???
Vo = 20.3 m/s @ 32 Deg.
Xo = 20.3*cos32 = 17.2 m/s.
Yo = 20.3*sin32 = 10.76 m/s.
a. hmax = 1.1 + (Yf^2-Yo^2/2g,
hmax=1.1 + (0-(10.76)^2) / -19.6=7.0 m.
b. Yf = 0 @ hmax.