You throw a ball from a cliff with an initial velocity of 15.0 m/sat an angle of$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$below the horizontal. Find (a) its horizontal displacement and (b) its vertical displacement 2.30 slater.

• Initial velocity of the ball,$V_i=15.0\mathrm{m}/\mathrm{s}$
• Angle of the cliff, $\theta =20.0°$.
• Time to be taken, $t=2.30\mathrm{s}$.

When the particle is launched with a certain velocity from the ground, the path of the particle can be calculated using the kinematic equations. The motion of the particle is called projectile motion. During this motion, the particle is acted upon by gravitational acceleration in the vertical direction. There is no acceleration in the horizontal direction. For calculation purposes, the air resistance is neglected and upward is taken to be a positive direction.

We can find horizontal displacement and vertical displacement using kinematic equations.

We are given the initial velocity and angle of projection of the ball. We can use horizontal and vertical components of initial velocity to find displacements.

Formulae:

The second kinematic equation of motion, $∆x=V_{0}t+\left(\frac{1}{2}\right)\times a\times t^2$ (i)

Here,$∆x$ the is displacement, $t$is time,$a$ is acceleration, $V_0$and is initial velocity.

We can write components of initial velocity

$$\begin{gathered} V_{ix}=V_i\cos \theta \\ {V}_{iy}=V_i\sin \theta \end{gathered}$$

Using the above equation and horizontal initial velocity component we can find displacement using equation (i). Substitute the given values in equation (i).

role="math" localid="1657014040430" $$\begin{gathered} d_x=V_i\cos +\left(\frac{1}{2}\right)a{t}^2 \\ =15\mathrm{m}/\mathrm{s}·\cos \left(-20°\right)\left(2.30\mathrm{s}\right)+\left(\frac{1}{2}\right)\left(0\right){\left(2.30\mathrm{s}\right)}^2 \\ =32.4\mathrm{m} \end{gathered}$$

Hence, the horizontal displacement of the ball is $32.4\mathrm{m}$.

Using the above equation and horizontal initial velocity component we can find displacement using equation (i) as follows. Substitute the given values in equation (i).

$$\begin{gathered} d_x=V_i\sin \theta ·t+\left(\frac{1}{2}\right)a{t}^2 \\ =15\mathrm{m}/\mathrm{s}·\sin \left(-20°\right)\left(2.30\mathrm{s}\right)+\left(\frac{1}{2}\right)\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(2.30\mathrm{s}\right)}^2 \\ =-37.7\mathrm{m} \end{gathered}$$

Hence, the vertical displacement of the ball is $37.7\mathrm{m}$. The sign is negative means in downward direction.