Question

a stone is thrown horizontally at a speed of 10m/s from the top of a cliff 80m high;
a,how long does it take for the stone to reach the bottom of the cliff?
b,how far from the base of the cliff does the stone strike the round?

BEST ANSWER + EXPLANATION WILL BE MARKED BRAINLIEST& 5 STARS

Answer 1

Given:-

• Height of the cliff = $\sf H = 80\ m$
• As the stone is thrown horizontally, it means vertical component of velocity if 0 and the initial velocity of 10 m/s is provided by the horizontal component of velocity alone. So, $\sf u_y = 0\:m/s^2$ and $\sf u_x = 10\:m/s^2$

Formula to be used:-

• $\sf s = ut + \dfrac{1}{2}at^2$ where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

Answer (a):- Finding time:-

In y direction:-

$\sf s_y = u_yt + \dfrac{1}{2}a_yt^2$

We know that $\sf u_y = 0\:m/s$. And putting $\sf s_y = - H, a_y = -g$ as they are acting in downward direction,

$\sf \longrightarrow - H = \dfrac{1}{2}\times (-g)\times t^2$

$\sf \longrightarrow \: - 80 = - \dfrac{1}{2} \times 10 \times {t}^{2}$

$\sf \longrightarrow \dfrac{80 \times 2}{10 } = {t}^{2}$

$\sf \longrightarrow t^2 = 16$

$\sf \longrightarrow t = \pm 4$

Since time cannot be negative,

$\longrightarrow \underline{\underline{\sf{\green{ t = 4\: seconds }}}}$

Answer (b):- Finding R:-

Let the distance between base of cliff and the point where stone strikes the ground be $\sf R.$

In x direction:-

$\sf s_x = u_xt + \dfrac{1}{2}a_xt^2$

Since there is no acceleration in x direction, $\sf a_x = 0\:m/s^2$

$\sf \longrightarrow R = u_xt$

Putting $\sf u_x = 10\:m/s$ and $\sf t = 4\:s$

$\sf \longrightarrow R = 10\times 4$

$\longrightarrow \underline{\underline{\sf{\green{R = 40\:meter}}}}$

Other formulae:-

• $\sf v = u + at$ where v is final velocity, u is initial velocity, a is acceleration and t is time taken.
• $\sf v^2 = u^2 + 2as$ where v is final velocity, u is initial velocity, a is acceleration and s is displacement.
• $\sf s_{n^{th}} = u + \dfrac{1}{2}a(2n - 1)$ where $\sf s_{n^{th}}$ is the displacement in $\sf n^{th}$ second.
• Equation of trajectory (only for ground to ground projectile):- $\sf y = x\:tan(\theta) - \dfrac{gx^2}{2u^2\:cos^2\theta}$ where $\theta$ is the angle of projection.