Question

Explore
A baseball is thrown from the top of a tall building with an initial velocity of 10 m/s from a height of h= 11.5 m above the ground. Find its speed when it reaches the ground (a) if its launch angle is 37°, and (b) if it is launched horizontally.

Conceptualize
As the baseball moves upward and to the right, the force of gravity accelerates it downward. The vertical component of its velocity decreases in magnitude until the baseball reaches maximum height, after which it accelerates downward to the ground, all while moving to the right throughout.

Categorize
We ignore the size and shape of the ball and consider it to be a point mass, and we ignore air resistance. This is a problem in projectile motion, except that we can now find the speed from the kinetic energy, instead of applying Newton's laws directly.

(A) The Final Speed for a Launch Angle of 37°Analyze As the ball first moves upward, its potential energy increases while the magnitude of its vertical velocity decreases, decreasing its kinetic energy. Afterward, during the downward part of its motion, its potential energy decreases, as the speed of its downward motion increases, with increasing kinetic energy. Throughout the motion of the ball, its total mechanical energy:

E_mech = K + U = ½mv₀² + mgh

remains constant, and therefore equal to its initial value:

E_mech = ½mv₀² + mgh

When the ball reaches the ground at height 0, the total mechanical energy is:

E_mech = ½mv_f²

Equating the two expressions for E_mech gives:

v_f = √ v₀² + 2gh = √(10 m/s)² + 2(9.8 m/s²)(11.5 m) =  m/s

(B) The Final Speed for a Horizontal LaunchAnalyze Because the total mechanical energy does not change, the change in kinetic energy depends only on the change in potential energy, which depends only on the difference of initial and final height, rather than on the detailed path in between. Launching the ball horizontally at the given speed instead of vertically leaves the difference between final and initial heights unchanged, so that the speed upon reaching the ground is

m/s.

Finalize Although changing the launch angle for a given launch speed does not change the speed of the ball when it reaches the ground, can it affect the magnitude of the vertical component of velocity when the ball strikes the ground? If so, what launch result in the ball striking the ground with the strongest vertical component of velocity?

Answer 1

a)

vi=10m/s, h=11.5m

By law of conservation of energy,

MEi = MEf

KEi + PEi = KEf + PEf

1/2mvi^2 + mgh = 1/2mvf^2

1/2vi^2 + gh = 1/2vf^2

vi^2 + 2gh = vf^2

vf = sqrt[vi^2 + 2gh]

vf = sqrt[10^2 +2*9.811.5]

vf = 18 m/s

b)

Here we again law of conservation of energy and again we get vf= 18m/s

Means final speed depends on initial velocity and height of the object and not on angle of initial velocity