Question

A golf ball is initially on a tee when it is

struck by a golfer. The ball is given an
initial velocity of 50 m/s at a 37° angle. The
ball hits the side of a building that is 200
meters horizontally away from the golfer.
(a) What are the horizontal and vertical
components of the ball's initial
velocity?
(b) How much time elapses before the
ball strikes the side of the building?
(c) How far from the ground does the ball
strike the building?

Answer 1

a.

`horizontal=39.9` m/s

`vertical=30.1` m/s

b.

`t=5.009`

c.

`y=27.7`

Step-by-step explanation:

Lets write down what we were given.

Angle = 37°

Initial Velocity = 50 m/s

Displacement in x direction = 200 m

Take note:

I am having some trouble with the theta symbol so let theta =
`N`

Lets do question C first.

We know that time is equal to
`(displacement)/(velocity)` aka
`t=(x)/(v)`.

`x=v`₀ₓ
`t` ⇒
`(x)/(v_(0x) )` ⇒
`(x)/(v_(0) *cos(N))`

Now substitute the expression for t into the equation for the position.

`y=(v_(0)sin(N))*((x)/(v_(0)cos(N) ))-(1)/(2)g((x)/(v_(0)cos(N) )) ^(2)`

Rearranging terms, we have

`y=(tan(N)*x)-[(g)/(2(v_(0)cos(N))^(2) ) ]x^(2)`

Now lets substitute our numbers in for the variables. Then simplify.

`y=(tan37*200)-[(9.81)/(2(50*cos37)^(2) ) ]200^(2)`

`y=150.7108-[(9.81)/(2(50*cos37)^(2) ) ]200^(2)`

`y=150.7108-[0.0030761]200^(2)`

`y=150.7108-(0.0030761*40000)`

`y=150.7108-123.0444`

`y=27.7`

Now lets do question B.

Lets steal this from the last question.

We know that time is equal to
`(displacement)/(velocity)` aka
`t=(x)/(v)`.

`x=v`₀ₓ
`t` ⇒
`(x)/(v_(0x) )` ⇒
`(x)/(v_(0) *cos(N))`

Now substitute the expression for t into the equation for the position.

`y=(v_(0)sin(N))*((x)/(v_(0)cos(N) ))-(1)/(2)g((x)/(v_(0)cos(N) )) ^(2)`

We can substitute
`t` for
`(x)/(v_(0)cos(N) )`

`y=(v_(0)sin(N))*(t)-(1)/(2)g(t) ^(2)`

We can rewrite the equation as

`(v_(0)sin(N)(t)-(1)/(2)*(g(t)^(2))=y`

Now lets substitute our numbers in for the variables.

`(50sin(37)(t)-(1)/(2)*(9.81(t)^(2))=27.7`

After some painful algebra and factoring we get

`30.09075115t-4.905t^(2)=27.6664`

Subtract
`27.6664` from both sides.

`30.09075115t-4.905t^(2)-27.6664=0`

Use the quadratic formula to find the solutions.

`\frac{-b+-\sqrt{b^(2)-4ac } }{2a}`

After some more painful algebra we get

`t=5.00854263, 1.12616708`

1.126 does not make any sense so.

`t=5.009`

Finally lets do question A.

Lets draw a triangle. We have the velocity which is the hypotenuse and we have the angle. From there we can solve for the opposite and adjacent sides.

Let
`A=horizontal` and
`O=vertical`

`cos(37)=(A)/(50)`

`A=39.9`

`sin37=(O)/(50)`

`O=30.1`