91Ó°ÊÓ

The velocity contains two components, one is a horizontal component and another one is vertical.

According to newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, s, u, t, a, are displacement, initial velocity, time, and acceleration.

For the vertical and horizontal motion, the velocity component will be $u\sin \mathrm{θ}\text{and cos}\mathrm{θ}$respectively.

Angle = $53°$

The wideness of the river = $\text{40m}$

Far bank deepness = role="math" localid="1659370924174" $\text{15m}$

The river below the ramp =$\text{100m}$

From the horizontal motion, gravity for horizontal motion will be zero.

$$\begin{gathered} s=u\cos \mathrm{θ}t+\frac{1}{2}a{t}^2 \\ 40\text{m}=u\cos 53°t \\ ut=66.47\text{m} \end{gathered}$$

From the vertical motion,

$$\begin{gathered} s=u\sin \mathrm{θ}t+\frac{1}{2}a{t}^2 \\ \text{- 15.0m}=66.47\text{m}×\text{sin}53°+\frac{1}{2}×-9.8m/s^2×t^2 \\ t=3.727\text{s} \end{gathered}$$

Now the speed will be,

$$\begin{gathered} ut=66.47\text{m} \\ u=\frac{66.47\text{m}}{3.727\text{s}} \\ u=17.8m/s \end{gathered}$$

$$\begin{gathered} u=\frac{17.8m/s}{2} \\ u=8.9m/s \end{gathered}$$

From the vertical motion,

$$\begin{gathered} s=u\sin \mathrm{θ}t+\frac{1}{2}a{t}^2 \\ \text{- 100.0m}=8.9m/s×\text{sin}53°t+\frac{1}{2}×-9.8m/s^2×t^2 \\ t=5.30\text{s} \end{gathered}$$

From the horizontal motion, gravity for horizontal motion will be zero.

$$\begin{gathered} s=u\cos \mathrm{θ}t+\frac{1}{2}a{t}^2 \\ s=8.9m/s×\cos 53°×5.30\text{s} \\ s=28.4\text{m} \end{gathered}$$

The Speed at the top of the ramp will be $u=17.8m/s$and the landing distance will be $s=28.4\text{m}$.