baseball is thrown from the top of a tall building with an initial velocity of 1

Question

baseball is thrown from the top of a tall building with an initial velocity of 12.50 m/s from a height of h = 25.50 m above the ground.

(A) Find its speed when it reaches the ground if its launch angle is +32°. (In other words the launch angle is 32° above the horizontal.)

(B) Find its speed when it reaches the ground if it is launched horizontally.

CONCEPTUALIZE
As the baseball moves upward and to the right, the force of gravity accelerates it downward. For positive launch angles, the vertical component of its velocity decreases in magnitude until the baseball reaches maximum height, after which it increases, due to the constant gravitational acceleration downward. The horizontal motion is to the right throughout the baseball's flight.

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.

ANALYZE
(A) Find the final speed for a launch angle of +32°.

For positive launch angles, 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, and this increase in speed corresponds to an increase in kinetic energy. Throughout the motion of the ball, its total mechanical energy
Emech = K + U =
1
2
mv02 + mgh
remains constant, and therefore equal to its initial value:
Emech =
1
2
mv02 + mgh.
When the ball reaches the ground at height 0, the total mechanical energy is
Emech =
1
2
mvf2.
Equating the two expressions for Emech gives
vf =

v02 + 2gh
=

(12.50 m/s)2 + 2(9.8 m/s2)(25.50 m)
=
25.61

Correct: Your answer is correct.
m/s.

(B) Find the final speed for a horizontal launch.

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 _______

Explanation / Answer

a) components in y-direction = 12.5 * sin 32 = 6.62 m/s

speed = sqrt [6.622 + (2*9.8*25.5)] = 23.3 m/s

speed [x direction] = 12.5 * cos 32 = 10.6 m/s

speednet = sqrt[10.62 + 23.32]

speed when it reaches the ground = 25.6 m/s

b)  if it is launched horizontally

velocity is same

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