A person who is 1.1 m tall throws a ball with a speed of 5.2 m/s at an angle of 56 degrees above the horizontal. What are the x and y components of the velocity (in m/s) of the ball the instant before it strikes the ground? The acceleration due to gravity is 10 m/s2

Answer 1

Vo = 5.2m/s[56o].
Xo = 5.2*Cos56 = 2.91 m/s.
Yo = 5.2*sin56 = 4.31 m/s.

h=-Yo^2/2g + 1.1=-(4.31)^2/-20 + 1.1
= 2.03 m. Above gnd.

Y^2 = Yo^2 + 2g*h = 0 + 20*2.03 = 40.6
Y = 6.37 m/s.

X = Xo = 2.91 m/s.