Numerical Problems Based on Projectile Fired at an Angle with the Horizontal for Class 11 Physics

Given: $u = 15 \text{ ms}^{-1}$, $\theta = 37^\circ$, and take $g = 9.8 \text{ ms}^{-2}$. Using $\sin 37^\circ \approx 0.6018$ and $\cos 37^\circ \approx 0.7986$.

(i) Time to highest point ($t$):

(ii) Maximum height ($H$):

(iii) Horizontal Range ($R$):

(Note: If the textbook answer is 21.2 m, it implies $g \approx 10 \text{ ms}^{-2}$ and exact $3/5$ and $4/5$ ratios are used, yielding $R = 21.6 \text{ m}$, or a specific rounded value of $\sin 74^\circ$ was used).

(iv) Total time of flight ($T$):

Given: $u = 20 \text{ ms}^{-1}$, $\theta = 60^\circ$, $t = 0.5 \text{ s}$, $g = 9.8 \text{ ms}^{-2}$.

(i) Position ($x, y$) after 0.5 s:

Horizontal distance ($x$):

Vertical distance ($y$):

(ii) Velocity after 0.5 s:

Resultant velocity magnitude:

Direction ($\beta$ with horizontal):

Given: Maximum height $H = 10 \text{ m}$, initial velocity $u = 28 \text{ ms}^{-1}$.

The formula for maximum height is:

Therefore, the angle of projection is $\theta = 30^\circ$.

Given: $u = 140 \text{ ms}^{-1}$, horizontal range $R = 1 \text{ km} = 1000 \text{ m}$.

The formula for horizontal range is:

Since $\sin(30^\circ) = 0.5$, we have:

Range at $15^\circ$ is $R_{15^\circ} = 6 \text{ km}$.

Since $R = \frac{u^2\sin(2\theta)}{g}$, for $\theta = 15^\circ$:

The maximum horizontal range $R_{max}$ occurs when $\theta = 45^\circ$, which gives $\sin(90^\circ) = 1$.

Since the maximum possible range is 12 km, hitting a target 10 km away is absolutely possible by selecting the appropriate angle between 15° and 45°.

The maximum horizontal distance is achieved when the angle of projection is $45^\circ$.

To throw the ball to the maximum vertical height, the cricketer must throw it straight up ($\theta = 90^\circ$).

The maximum vertical height $H_{max}$ is:

Substitute $R_{max}$ into the equation:

Given: $u = 20 \text{ ms}^{-1}$, $\theta = 45^\circ$, Receiver’s initial distance $= 25 \text{ m}$, $g = 9.8 \text{ ms}^{-2}$.

First, find the total range ($R$) of the kick:

The distance the receiver needs to run is the difference between the ball’s landing spot and his starting position:

Now, calculate the time of flight ($T$) of the football:

The receiver’s required speed ($v_r$) is distance divided by time:

Notice the angle is given with the vertical ($60^\circ$). The angle with the horizontal is $\theta_1 = 90^\circ – 60^\circ = 30^\circ$.
Range $R_1 = 2 \text{ km}$.

Now, calculate the new range $R_2$ when fired at $45^\circ$ with the horizontal:

Substitute the value of $\frac{u^2}{g}$:

Given: Coordinates of the bird $(x, y) = (59.2 \text{ m}, 39.6 \text{ m})$, projection angle $\theta = 45^\circ$. Take $g = 9.8 \text{ ms}^{-2}$.

Use the equation of the trajectory of a projectile:

Substitute the given values:

Since $\tan 45^\circ = 1$ and $\cos^2 45^\circ = 0.5$:

Rearrange to solve for $u^2$:

Initial velocity $u = 10 \text{ ms}^{-1}$, $\theta = 30^\circ$ (above horizontal), Range $x = 17.3 \text{ m}$, $g = 10 \text{ ms}^{-2}$.

First, find the time of flight ($t$) using the horizontal motion equation:

Now, find the height of the tower ($h$) using the vertical motion equation. Let upward be positive, and take origin at the top of the tower. Final vertical displacement is $-h$.

Therefore, the height of the tower is $10 \text{ m}$.