Answer
(a) The car should be traveling at a speed of 24.1 m/s
(b) The speed of the car just before it lands is 31.0 m/s
Work Step by Step
(a) The car falls from a height of 21.3 m to a height of 1.8 m. The vertical distance is 21.3 - 1.8 m which is 19.5 m.
We can find the time it takes to fall freely.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{(2)(19.5~m)}{9.80~m/s^2}}$
$t = 1.99~s$
We can find the speed required to cross the 48.0-meter wide river in a time of 1.99 seconds.
$v_x = \frac{x}{t} = \frac{48.0~m}{1.99~s} = 24.1~m/s$
The car should be traveling at a speed of 24.1 m/s.
(b) $v_x = 24.1~m/s$
$v_y = \sqrt{2ay} = \sqrt{(2)(9.80~m/s^2)(19.5~m)}$
$v_y = 19.5~m/s$
We can find the speed of the car.
$v = \sqrt{v_x^2+v_y^2}$
$v = \sqrt{(24.1~m/s)^2+(19.5~m/s)^2}$
$v = 31.0~m/s$
The speed of the car just before it lands is 31.0 m/s
