Question

Question 11 : A baseball with a mass of 142.5 g is thrown from the roof of a building 50 m above the street. The ball is thrown with an initial speed of 28 m/s and at an angle of 60◦ with respect to horizontal.

A) What are the initial kinetic energy, potential energy, and velocity components of the ball?

B) What are the components of the ball’s velocity vector at the apex of its flight path?

C) What is the kinetic energy of the ball at the apex (Hint: It is not zero.)

D) What is the elevation of ball relative to the top of the building at the apex?

E) What are the Kinetic Energy and speed of the ball just before it hits the street?

Answer 1

a)

Initial Kinetic energy

K_i=(1/2)mV² =(1/2)*0.1425*28² =55.86 J

Initial Potential energy

U_i=mgh=0.1425*9.81*50 =69.9 J

Initial Horizontal and vertical components of velocities are

V_ox=28Cos60=14 m/s

V_oy=28Sin60=24.25 m/s

b)

at the maximum height ,final velocity in vertical direction is zero.

V_fy=0

V_fx=V_ox =14 m/s

therefore

V_f=(14 i ) m/s

c)

K=(1/2)mV²=(1/2)(0.1425)(14)² =13.965 J

d)

From

V_fy²=V_oy²-2g(H-h)

=>0=24.25²-2*9.81*(H-50)

H=79.97 m=80 m

e)

From

Y=Y_o+V_oyt-(1/2)gt²

when the ball hits the street Y=0

0=50+(24.25)t-(1/2)(9.81)t²

4.905t²-24.25t-50=0

t=6.51

From

V_fy=V_oy-gt =24.25-9.81*6.51

V_fy =-39.61 m/s

V_fx=V_ox =14 m/s

final speed of the ball when it hits the street

V_f=sqrt[14²+(-39.61)²] =42.01 m/s

Kinetic energy

K=(1/2)(0.1425)(42.01)² =125.76 Joules