Question

A ball is projected with a certain speed from a point on the ground which is at a distance of 30 m from a vertical wall. If the angle of projection is 45∘ with the horizontal, the ball just clears the top of the wall and strikes the ground at a distance of 10 m from the wall on the other side. The height of wall is h. Find 10h. (Take g=10 ms−2

Answer 1

75

Answer 2

Solution

1. Total Range:

For a projectile with initial speed $u$ and angle $45^\circ$, the range is

$$R=\frac{u^2\sin 90^\circ}{g}=\frac{u^2}{g}$$

Given that the ball lands 10 m beyond the wall which is 30 m from the launch point,

$$R=30+10=40\text{ m}$$

Thus,

$$40=\frac{u^2}{10}\quad\Rightarrow\quad u^2=400\quad\Rightarrow\quad u=20\text{ m/s}$$

2. Height at the Wall (x = 30 m):

Time to reach the wall:

$$t=\frac{x}{u\cos 45^\circ}=\frac{30}{20\cdot\frac{1}{\sqrt{2}}}=\frac{30\sqrt{2}}{20}=\frac{3\sqrt{2}}{2}\text{ s}$$

Vertical height at time $t$:

$$y= u\sin 45^\circ\cdot t -\frac{1}{2}gt^2 = \frac{20}{\sqrt{2}}\cdot \frac{3\sqrt{2}}{2} - 5\left(\frac{3\sqrt{2}}{2}\right)^2$$

Simplify:

$$y= \frac{20\cdot 3}{2} - 5\left(\frac{18}{4}\right)=30 - 5\cdot4.5=30-22.5=7.5\text{ m}$$

So the wall height $h=7.5\text{ m}$.

3. Final Answer:

Since the problem asks for $10h$, we have:

$$10h=10\times7.5=75$$

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Explanation:

• Calculated range to get initial speed.
• Computed time to wall and substituted in the vertical displacement equation to get the height at 30 m.
• Found $h$ and then $10h$.