91Ó°ÊÓ

The time of flight refers to the time period for which a projectile remains in the air.

The initial velocity of the rock,$v_0=15\mathrm{m}/\mathrm{s}$

Let us assume that at the point of landing, the horizontal displacement is x, the vertical displacement is y, and the time taken by the rock to hit the slope is t.

The initial velocity of the rock in the horizontal direction,$v_{\mathrm{x}}=v_0$

Using the second equation of motion for the horizontal motion, you can write:

$\begin{array}{l}x=v_{\mathrm{x}}t+0\\ x=v_{0}t\end{array}$

(As acceleration in the horizontal direction is zero)

Substituting the values in the above equation, you get:

role="math" localid="1644994977727" $x=15t…\left(i\right)$

The initial velocity along the vertical direction will be:

$v_{\mathrm{y}}=0$(The projectile has been launched horizontally)

Applying the second equation of motion for the vertical direction, you get:

$y=0+\frac{1}{2}g{t}^2$

Substituting the values in the above equation, you get:

$y=0+\frac{1}{2}×\left(-9.8\right)×t^2$

The above equation can be rewritten as:

role="math" localid="1644995045620" $y=-4.9t^2…\left(ii\right)$

As the angle of slope is $45°$, and the value of $\tan 45°=1$, the magnitude of displacement in the vertical direction is equal to the displacement in the horizontal direction.

$\left|x\right|=\left|y\right|$

Substituting the values from equations (i) and (ii), you get:

$15t=4.9t^2$

Solving the above equation for t (ignoring solution t = 0 s), you get:

t = 3.06 s

Thus, the rock will hit the slope at time t = 3.06 s.