Question

A girl stands on top of the Golden Gate Bridge and throws a penny at a speed of 94.32km/hr at an angle of -68.8°. How long does it take the penny to fall the 224m to the San Francisco Bay?

How far from the base of the bridge does it land?
What is the penny's velocity when it enters the water and at what angle?

Answer 1

Now initial velocity of projection `=94.32\ kmph` `=94.32xx1000/3600=26.2\ ms^-1`
To calculate the time taken by the penny to fall to the bay we need to know its vertical velocity.

Which is `=26.2 sin (-68.8)=>-26.2 sin (68.8)\ ms^-1` (in the downwards direction.)
Applicable kinetic expression in the vertical direction is

`h = ut + 1/2 g t^2`

Taking origin at the location of girl and inserting given values and remembering that acceleration due to gravity `g=9.81\ ms^-2` acts in the downwards direction, we get

`-224 = -26.2 sin (68.8^@)t + 1/2 (-9.81) t^2`
`=> t^2+4.98t -45.67=0`

Solving above quadratic using builtin graphics utility and ignoring the `-ve` root as time can not be negative we get

`t=4.7\ s`, rounded to one decimal place.

The distance from the base of the bridge does it lands is given by the equation

`R="Horizontal Velocity"xx"time taken"`
`R=(26.2cos(-68.8^@))xx4.712=44.64\ m`

There is no change in the horizontal component of penny's velocity when it enters the water. Vertical component is found using the following kinematic expression.

`v=u+g t`

We get

`v_v=-26.2 sin (68.8)+-(9.81) xx4.712`
`=>v_v=-24.427-43.282`
`=>v_v=-67.709\ ms^-1`

Penny's velocity when it enters water `v_f=-67.709hati+9.474hatj`

Magnitude of the final velocity `=sqrt((-67.709)^2+(9.474)^2)`
`=68.368\ ms^-1`

Angle of this velocity can be calculated with the help of `tanalpha` formula , where `alpha` is the angle between the resultant and one of the vectors. It is left to the student to complete.