Question

The girl's brother stands at the edge of the Golden Gate Bridge and throws a penny upwards at 29 m/s at an angle of 34°. How long does it take the penny to fall to the water, 227m below? How far from the base of the bridge does the penny hit the water?

Answer 1

Total time of flight is 8,66 s, and range of projectile is 2082 m.

Explanation:

I will break the projectile motion of the penny up into different stages :
A: Release
B : Max height
C: Release height.
D: Water.
In addition, I will consider horizontal (X) and vertical (Y) directions separately for each stage.
The X-Direction:
In the x-direction, there is no acceleration and hence constant velocity of `29cos34^@=24,042m//s`
The Y-Direction:
In the y-direction, there is constant acceleration throughout due to gravity of value `g=-9,8m//s^2`

Now we get:

Maximum height is reached at point B, so working from A - B in the y-direction we get `s=ut+1/2at^2 =13,417m`
Hence the maximum height above the water is `227+13,417=240,417m`

Since it reaches its maximum height after 1,655s, its horizontal range at this time in the x-direction is `s_x =vcosθ⋅t=39,79m`

Thus, by Pythagoras, the distance from the bridge at this point is `sqrt(39,79^2 +13.417^2)=42m`