Question

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.5 m above the river, whereas the opposite side is a mere 1.5 m above the river. The river itself is a raging torrent 69.0 m wide.
Required:
a. How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?
b. What is the speed of the car just before it lands safely on the other side?

Answer 1

Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

$\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y = \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} } =2\ m/s$

Also:

$\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x = u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s$

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

$v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s$