91Ó°ÊÓ

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,

$d_x=v_0\cos \mathrm{θ}xt$

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

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

It is given that, for vertical motion,

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

Again using equation (ii)

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

$d_x=27.6m$

Using equation (i),

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

$d_y=7.26\mathrm{m}$

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

$d_y=0$

Thus,

$v_0\sin \mathrm{θ}×t-\frac{1}{2}g{t}^2=0$

$t=\frac{2v_0}{g}\sin \mathrm{θ}$

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

$t=2.62s$

For this time x component is,

$d_x=v_0\cos \mathrm{θ}xt$

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

$d_x=40.2m$

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