A man stands on the roof of a building of height 13.0m and throws a rock with a velocity of magnitude 33.0m/s at an angle of 25.3(degree) above the horizontal. You can ignore air resistance. (A)Calculate the maximum height above the roof reached by the rock. (B)Calculate the magnitude of the velocity of the rock just before it strikes the ground.(C)Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

(A)
The vertical component of velocity is,
$V_y=V_0\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$
Substitute 33.0 m/s for $v_0\text{ and }25.3\circ <!--\; \circ \; -->for\mathrm{\theta <!--\; \theta \; -->}$
$v_y=(33.0m/s)\sin <!--\; \; -->(25.3\circ <!--\; \circ \; -->)=14.1m/s$
Calculate the maximum height above the roof reached by the rock by using the expression of the maximum height.
The maximum height reached by the projectile is,
$h_{max}=\frac{(v_0\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2}{2g}$
Substitute 14.1 m/s for $v_0\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\text{ and }9.8m/s^{2}forg.$
$h_{max}=\frac{2(9.8m/s^2)}{(14.1m/s{)}^2}=10.14m$
The maximum height is 10.14 m.
(B)
The total height reached by the rock.
The height is,
$H=h_0+h_{max}$
Here, H is the total height, $h_0$ is the initial height, and $h_{max}$ is the maximum height.
Substitute 13.0 m for $h_0$ and 10.14 m for $h_{max}$
$$\begin{gathered} H=13.0m+10.14m \\ =23.14m \end{gathered}$$
The total height can be calculated by the summation of the initial height of the building and the maximum height from the roof of the building.
Calculate the velocity in the vertical direction just before the rock hit the ground.
The velocity is,
$v_y{}^2=u_y{}^2+2as$
The vertical component of velocity at the maximum height will be zero.
Substitute 0 m/s for $u_y$ ,g for a, and H for s.
$$\begin{gathered} v_y{}^2=(0m/s{)}^2+2as \\ {v}_y{}^2=2gH \end{gathered}$$
Substitute 23.14 m for H and 9.8m/s${}^2$ for g
$$\begin{gathered} v_y{}^2=2(9.8m/s^2)(23.14m) \\ {v}_y=\sqrt{2(9.8m/s^2)(23.14m)} \\ {v}_y=21.3m/s \end{gathered}$$
The horizontal component of the velocity will remain unchanged before the rock hit the ground.
The horizontal component of velocity is,
$v_x=v_0\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$
Substitute 33.0 m/s for $v_0\text{ and }25.3\circ <!--\; \circ \; -->\text{ for }\mathrm{\theta <!--\; \theta \; -->}$
$v_x=(33.0m/s)\cos <!--\; \; -->(25.3\circ <!--\; \circ \; -->)v_x=29.83m/s$
The magnitude of the velocity is,
$\mid <!--\; \mid \; -->v\mid <!--\; \mid \; -->=v_x{}^2+v_y{}^2$
Substitute 29.83 m/s for $v_x$ and 21.3 m/s $v_y$
The magnitude of the velocity is 36.7 m/s.
$\mid <!--\; \mid \; -->v\mid <!--\; \mid \; -->=\sqrt{(29.83m/s{)}^2+(21.3m/s{)}^2}=36.7m/s$
(C)
The time of flight $t_1$ when the rock reaches at the maximum height is,
$t_1=\frac{v_0\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{a}$
Substitute 33.0 m/s for $v_0,9.8m/s^2\text{ for }a,\text{ and }25.3\circ <!--\; \circ \; -->for\mathrm{\theta <!--\; \theta \; -->}$
$t_1=\frac{(33.0m/s)sin(25.3\circ <!--\; \circ \; -->)}{9.8m/s^2}=1.44s$
The time of flight $t_2$ from the maximum height to the point where the rock hit the ground is,
$t_2=\frac{v_y-u_y}{a}$
The vertical component of the velocity at the maximum height is zero.
Substitute 21.3 m/s for $v_y,0m/s\text{ for }{u}_y,\text{ and }9.8m/s^2\text{ for a.}$
$t_2=\frac{(21.3m/s)-(0m/s)}{9.8m/s^2}=2.17s$
The total time T of the flight is,
$T=t_1+t_2$
Substitute 1.44 s for $t_1\text{ and }2.17s\text{ for }{t}_2$
T=(1.44s)+(2.17s)=3.61s
The total time of flight is 3.61 s.
Determine the horizontal distance.
The horizontal distance is, $s=u_{x}t$
Substitute 29.83 m/s for $u_x\text{and }3.61s\text{ for }t.$
$s=(29.83m/s)(3.61s)=107.69m$
The horizontal distance is 107.69 m.

The concept used to solve the problem is kinematics.
first of all, calculate the maximum height above the roof reached by way of the rock. Later calculate the significance of the rate of the rock simply earlier than it strikes the ground. subsequently, calculate the horizontal distance from the bottom of the building to the factor where the rock moves the floor.
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The projectile motion can be defined as the motion of a particle when thrown, follows the curved path because of gravity.
The x-component and the y-component of the initial velocity is,