Question

A baseball thrown at an angle of 60.0° above the hori- zontal strikes a building 18.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance. a) Find the magnitude of the ball’s initial velocity (the velocity with which the ball is thrown). b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.

Answer 1

a)
`v_(o)=16.6m/s`

b)
`v=14.6m/s`

`\beta =-55.32º`

Step-by-step explanation:

From the exercise we know that the ball is thrown at an given angle, strikes the building 18.0 m away, that will be our horizontal displacement. Also, we know whats the final height of the ball which is 8.0 m

a) Given the information that we have we can calculate the initial velocity by solving the following formula

`x=v_(o)cos(60)t`

Since we don't know how much time does the ball take to strike the building we need to calculate that first

`v_(o)=(x)/(tcos(60))=(18m)/(tcos(60))` (1)

If we analyze the vertical displacement

`y=y_(o)+v_(oy)t+(1)/(2)gt^2`

`8m=(18m)/(tcos(60))sin(60)t-(1)/(2)(9.8m/s)t^2`

`(4.9m/s^2)t^2=(18m)tan(60)-8m`

`t=\sqrt{((18m)tan(60)-8m)/(4.9m/s^2) }=2.17s`

Knowing that time we can calculate the ball's initial velocity from (1)

`v_(o)=(18m)/((2.17s)cos(60)) =16.6m/s`

b) To calculate the magnitude and direction of the ball's velocity we need to find the x and y components of velocity

`v_(x)=v_(ox)+at=(16.6m/s)cos(60)=8.3m/s`

`v_(y)=v_(oy)+gt=(16.6m/s)sin(60)-(9.8m/s^2)(2.17s)=-12m/s`

The magnitude of velocity is:

`v=\sqrt{v_(y)^(2)+v_(x)^2}=√((-12m/s)^2+(8.3m/s)^2)=14.6m/s`

The direction of the ball's velocity is:

`\beta =tan^(-1)((v_(y) )/(v_(x)))=tan^(-1)((-12m/s)/(8.3m/s))=-55.32º`