Driving Off The Edge

In the vertical direction-- when the car just takes off from the cliff, it has got only initial horizontal velocity and zero initial vertical velocity. However, there is no acceleration in the horizontal direction and the acceleration in the vertical direction is acceleration due to gravity, g(10 m/s²). The distance covered in the horizontal direction is 160m and that in the vertical direction is 80m.

Vertical motion: We apply the equation of motion, h = ut + 1/2 gt². Here u = 0 and h = 80. Substituting we get, t = 4s.

Horizontal motion-- Since there is no acceleration in the horizontal direction, s = vt where s is the distance covered and v is velocity in the vertical direction. Substituting for time from above, we get v = s/t = 160/4 = 40 m/s.

we know that $a=\frac{2d}{t^2}$ where a=acceleration, d=distance,t=time incase $u=0$ . we get $t=\sqrt{\frac{2\times 80}{10}}$ since a=g= $10m/s^2$ $t=4 sec$ $v=\frac{d}{t}$ $v=\frac{160}{4}$ so $v=\boxed{40 m/s}$

For the motion of bodies in horizontal projection, as this case, is governed by some laws of motion from which we can derive certain relation of the Horizontal Distance covered $(R)$ and Vertical height $(h)$ , and the initial velocity $(u)$ , such that

$R=u \cdot \sqrt{\frac{2h}{g}}$

$\Rightarrow$ $160=u \cdot \sqrt{\frac{2 \cdot 80}{10}}$

$\Rightarrow$ $u=\boxed{Forty}$ $\boxed{ms^{-1}}$

Arya :)

since it went over the cliff horizontally with horizontal velocity. Vertical velocity is 0 . acclereration is -10 . displacement is -80. by using v^2=u^2 + 2as you can find out the initial velocity. ps , using sign convention

• Time to touch the ground:
• $80 = 5t^{2}$
• Therefore t = 4
• $u = \frac{m}{s} = \frac{160}{4} = 40$