91影视

It is given that, initial velocity has the magnitude 20.0 m/s and the projection of angle is$40掳$ 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鈥檚 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,

$d_x=v_0\cos 胃xt$

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

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

It is given that, for vertical motion,

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

Again using equation (ii)

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

$d_x=27.6m$

Using equation (i),

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

$d_y=7.26\mathrm{m}$

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

$d_y=0$

Thus,

$v_0\sin 胃脳t-\frac{1}{2}g{t}^2=0$

$t=\frac{2v_0}{g}\sin 胃$

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

$t=2.62s$

For this time x component is,

$d_x=v_0\cos 胃xt$

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

$d_x=40.2m$

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