Numerical Problems Based on Class 11 Physics Projectile Motion

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

To achieve the maximum vertical height, the ball must be thrown straight up ($\theta = 90^\circ$).

Position: Horizontal distance ($x$) and vertical drop ($y$) after $t = 3\text{ s}$:

Velocity: The horizontal velocity remains constant ($v_x = 50\text{ m/s}$). The vertical velocity is $v_y = gt = 9.8 \times 3 = 29.4\text{ m/s}$.

Given that Range ($R$) = $2 \times$ Maximum Height ($H$):

Expanding $\sin 2\theta$ to $2 \sin \theta \cos \theta$:

From a right triangle where opposite=2 and adjacent=1, hypotenuse=$\sqrt{5}$. Thus, $\sin \theta = \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$. Substituting back into the Range formula:

First, find the value of $\frac{u^2}{g}$ using the given range at $30^\circ$:

The maximum possible range occurs at an angle of $45^\circ$, which is equal to $R_{max} = \frac{u^2}{g}$.

Since the maximum possible range is approximately 3.46 km, it is physically impossible to hit a target 5.0 km away with this fixed muzzle speed.

Given $u = 19.6\text{ m/s}$, $\theta = 30^\circ$, and $g = 9.8\text{ m/s}^2$.

Time of Flight ($T$):

Horizontal Range ($R$):

To maximize range without hitting the ceiling, the maximum height of the projectile should exactly equal the ceiling height ($H = 25\text{ m}$). Using $g = 9.8\text{ m/s}^2$:

Now, calculate the horizontal range using this angle:

Note: For a launch angle of $45^\circ$, the trajectory equation $y = x \tan(45^\circ) – \frac{gx^2}{2u^2 \cos^2(45^\circ)}$ simplifies to $y = x – \text{something positive}$. This means the height $y$ MUST be less than the horizontal distance $x$. Since the provided text has height ($39.6$) > distance ($35.2$), it is physically impossible. We assume a common textbook typo where horizontal distance $x = 39.6\text{ m}$ and height $y = 35.2\text{ m}$.

To cross the 500m hill in minimum flight time, the packet must just clear the peak. This means the maximum height of the trajectory should be exactly 500m.

At this optimal angle ($\theta \approx 53.1^\circ$, $\cos \theta = 0.6$), the horizontal distance to the peak is exactly half the range:

Since the hill is 800m away, the cannon must be moved closer by $800 – 750 = 50\text{ m}$ to align the trajectory peak with the hill peak.

Total shortest time = $25 + 20 = 45\text{ s}$.

When a problem states an object “just passes through” an opening without specifying an angle, it is standard to assume the window is at the highest point of the trajectory.

So, $H_{max} = 19.6\text{ m}$ and half-range $R/2 = 39.2\text{ m} \implies R = 78.4\text{ m}$.

Dividing the two equations gives $\frac{\tan \theta}{4} = \frac{19.6}{78.4} = \frac{1}{4} \implies \tan \theta = 1 \implies \theta = 45^\circ$.

Substitute $\theta = 45^\circ$ back into the height equation ($g = 9.8\text{ m/s}^2$):

Yes, the bullet will hit the monkey (assuming the bullet reaches the tree before hitting the ground). This is because gravity acts on both the bullet and the monkey equally.

If there were no gravity, the bullet would travel in a straight line to the monkey’s initial position. With gravity, the bullet falls a distance of $\frac{1}{2}gt^2$ below that straight line. Simultaneously, the dropping monkey also falls exactly $\frac{1}{2}gt^2$ from its branch. Thus, they will always intersect in mid-air.