Question

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance. Part A
Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).
Express your answer in meters per second.
Part B
Find the magnitude of the velocity of the baseball just before it strikes the building.
Express your answer in meters per second.
Part C
Find the direction of the velocity of the baseball just before it strikes the building.
Express your answer in degrees.

Answer 1

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its initial velocity

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the magnitude of the ball just before it strikes the building we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The direction of the ball is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$