91影视

The given data can be listed below,

• Height of the building is, h = 15 m
• The velocity of rock is, v₀ = 30m/s
• The angle at which rock is thrown, $胃={33}^{鈭�}$

A projectile is any object or body that is launched at an angle to the horizontal, and its motion is referred to as projectile motion. Normally, the route is a parabola.

The horizontal component of acceleration is zero because of the constant velocity of the rock and vertical component of acceleration of rock is acceleration due to gravity in downward direction.

The maximum height attain by the rock is given by,

$h_{max}=\frac{{\left(v_0\sin 胃\right)}^2}{2g}$

Here, v₀ is the velocity of rock, g is the acceleration due to gravity.

Substitute all the values in the above,

$\begin{array}{rcl}{h}_{max}& =& \frac{{\left(30\sin {33}^o\right)}^2}{2\left(9.8m/s^2\right)}\\ & =& \frac{900m^2/s^2{\left(0.544\right)}^2}{19.6m/s^2}\\ & =& 13.6m\end{array}$

Thus, the maximum height the rock can reach is 13.6 m

The vertical component of velocity' at maximum height is zero. The initial horizontal component of velocity is given by,

$v_{0x}=v_0\cos 胃$

Here, v₀ is the velocity of rock.

Substitute all the values in the above,

$\begin{array}{rcl}{v}_{0x}& =& \left(30m/s\right)\cos {33}^o\\ & =& 25.2m/s\end{array}$

The initial vertical velocity component is given by,

$v_{iy}=v_0\sin {伪}_0$

The vertical component of velocity just before rock hits the ground is given by,

$\begin{array}{rcl}{v}_y^2& =& v_{0y}^2+2a_y鈭�y\\ & =& v_{0y}^2+2g\left(y_f-y_i\right)\\ v_y& =& 卤\sqrt{{\left(v_0\sin {伪}_0\right)}^2-2g\left(y_f-y_i\right)}\end{array}$

Substitute all the values in the above expression.

$\begin{array}{rcl}{v}_y& =& 卤\sqrt{{\left(30\sin {33}^o\right)}^2-2脳9.80\left(0-15\right)}\\ & =& 卤23.7m/s\end{array}$

The rock鈥檚 vertical velocity component must be negative just before it hits the ground. So,

$v_y=-23.7m/s$

The magnitude of the velocity of the rock just before it hits the ground is given by,

$v=\sqrt{v_x^2+v_y^2}$

Here, v_x is the velocity in horizontal direction and v_y is the velocity of rock in vertical direction.

Substitute all the values in the above expression.

$\begin{array}{rcl}v& =& \sqrt{{\left(25.2m/s\right)}^2+{\left(-23.7m/s\right)}^2}\\ & =& 34.6m/s\end{array}$

Thus, the velocity of rock just before hitting the ground is 34.6m/s.

The horizontal velocity component of the rock is constant, using the velocity kinematic formula the horizontal range from the base of the building to the point where the rock strikes the ground is given by,

$鈭�x=v_{0x}t+\frac{1}{2}{a}_x{t}^2$

Here, v_0x is the initial horizontal velocity component of the rock, t is the time of flight, and a_x is the horizontal component of the acceleration.

Substitute all the values in the above,

$\begin{array}{rcl}鈭�x& =& v_{ix}t+0\\ & =& v_0\cos {伪}_{0}t\end{array}$

The time taken by rock to strike ground is given by,

$\begin{array}{rcl}{v}_{fy}& =& v_{iy}+a_{y}t\\ v_{fy}& =& v_0\sin {伪}_0-gt\\ t& =& \frac{v_0\sin {伪}_0-v_{fy}}{g}\end{array}$

Substitute all the values in the above,

$\begin{array}{rcl}t& =& \frac{\left(30m/s\right)\sin {33}^o-\left(-23.7m/s\right)}{9.8m/s^2}\\ & =& 4.08s\end{array}$

The horizontal range is calculated by substituting all the values in the expression,

$\begin{array}{rcl}鈭�x& =& v_0\cos 胃t\\ & =& \left(4.08s\right)v_0\cos 胃\\ & =& \left(4.08s\right)\left(30m/s\right)\cos {33}^o\\ & =& 103m\end{array}$

Thus, the horizontal range of the rock from the base of building is 103m.

The motion for position in x-direction as a function of time is shown in the graph given below.

The motion for position in y-direction as a function of time is shown in the graph given below,

The velocity in x-direction as a function of time is shown in the graph given below,

The velocity in y-direction as a function of time is shown in the graph given below.