A stone is catapulted at time t = 0 , with an initial velocity of magnitude 20.0 m/s and at an angle of $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal. What are the magnitudes of the (a)horizontal and (b) vertical components of its displacementfrom the catapult site atat t=1.10 s Repeat for the (c) horizontal and (d) vertical components at t = 1.8 s , and for the (e)horizontal and (f) vertical components at $\mathit{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{s}$ .

It is given that, initial velocity has the magnitude 20.0 m/s and the projection of angle is$40°$ above the horizontal.

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Using the second kinematic equation, the magnitude of the horizontal and vertical components of the stone’s displacement from the catapult site at different times will be computed.

Formula:

The displacement in kinematic equation for vertical motion is given by,

$d_y=v_{0}t+\frac{1}{2}a{t}^2$

Displacement for horizontal motion is given by,

$d_x=v_0\cos \theta xt$

Using equation (ii) for horizontal motion and substituting the given values in equation (ii),

$$\begin{gathered} d_x=\left(20\frac{m}{5}\right)\cos 40°\times (1.10s) \\ {d}_x=16.9m \end{gathered}$$

It is given that, for vertical motion,

$$\begin{gathered} d_y=v_0\sin \theta \times t-\frac{1}{2}g{t}^2 \\ {d}_y=\left(20.0\frac{m}{s}\right)\sin 40°\times (10s)-\frac{1}{2}\left(9.8\frac{m}{s^2}\right){\left(1.10s\right)}^2 \\ {d}_y=8.21m \end{gathered}$$.

Again using equation (ii)

$d_x=\left(20.0\frac{m}{5}\right)\cos 40°\times \left(1.80s\right)$

$d_x=27.6m$

Using equation (i),

$d_y=\left(20.0\frac{m}{s}\right)\sin 40°\times (10s)-\frac{1}{2}\left(9.8\frac{m}{s^2}\right){\left(1.10s\right)}^2$

$d_y=7.26\mathrm{m}$

To find the time when stone hits the ground,it is given that,

$d_y=0$

Thus,

$v_0\sin \theta \times t-\frac{1}{2}g{t}^2=0$

$t=\frac{2v_0}{g}\sin \theta$

$t=\frac{2\left(20.0{\displaystyle \frac{m}{s}}\right)}{\left(9.8{\displaystyle \frac{m}{s^2}}\right)}\sin 40°$

$t=2.62s$

For this time x component is,

$d_x=v_0\cos \theta xt$

$d_x=\left(20.0\frac{m}{s}\right)\cos 40°\times \left(2.62s\right)$

$d_x=40.2m$

When stone lands at $t=5s$, vertical component of stone is $d_y=0\mathrm{m}$