Find the velocity and direction of projection of a shot which passes in a horizontal direction just over the top of a wall which is 250m off and 125m high. ( g=9.8ms^ 2 ) [6]

Solution

We analyze the motion of the projectile in two dimensions: horizontal (x) and vertical (y). The key is that the projectile just clears the wall, meaning it reaches the wall's height ($y=125\text{m}$) at the same horizontal distance ($x=250\text{m}$) with zero vertical velocity (since it is moving horizontally at the top).

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Step 1: Equations of Motion

The general equations for projectile motion (ignoring air resistance) are:

• Horizontal motion (constant velocity): $$x=u\cos \theta \cdot t$$

• Vertical motion (under gravity): $$y=u\sin \theta \cdot t-\frac{1}{2}g{t}^2$$ $$v_y=u\sin \theta -gt$$

At the wall's top:

• $x=250\text{m}$
• $y=125\text{m}$
• $v_y=0$ (horizontal motion)

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Step 2: Time to Reach the Wall

From the vertical motion equation at the wall's top ($v_y=0$): $$0=u\sin \theta -gt⟹t=\frac{u\sin \theta }{g}$$

Substitute $t$ into the vertical displacement equation: $$125=u\sin \theta \left(\frac{u\sin \theta }{g}\right)-\frac{1}{2}g{\left(\frac{u\sin \theta }{g}\right)}^2$$ Simplify: $$125=\frac{u^2{\sin }^2\theta }{g}-\frac{u^2{\sin }^2\theta }{2g}=\frac{u^2{\sin }^2\theta }{2g}$$ $$u^2{\sin }^2\theta =250g\text{(1)}$$

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Step 3: Horizontal Distance

From the horizontal motion equation: $$250=u\cos \theta \cdot t=u\cos \theta \left(\frac{u\sin \theta }{g}\right)=\frac{u^2\sin \theta \cos \theta }{g}$$ $$u^2\sin \theta \cos \theta =250g\text{(2)}$$

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Step 4: Solve for Angle $\theta$

Divide equation (2) by equation (1): $$\frac{u^2\sin \theta \cos \theta }{u^2{\sin }^2\theta }=\frac{250g}{250g}⟹\cot \theta =1⟹\theta =45^{\circ }$$

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Step 5: Calculate Initial Velocity $u$

Substitute $\theta =45^{\circ }$ into equation (1): $$u^2{\sin }^{2}45^{\circ }=250\times 9.8$$ $$u^2{\left(\frac{1}{\sqrt{2}}\right)}^2=2450⟹u^2\cdot \frac{1}{2}=2450$$ $$u^2=4900⟹u=70\text{m/s}$$

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Step 6: Velocity and Direction at Projection

• Velocity of projection ($u$): $70\text{m/s}$
• Direction of projection ($\theta$): $45^{\circ }$ to the horizontal

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Verification

Check if the projectile just clears the wall:

• Time to reach the wall: $$t=\frac{u\sin \theta }{g}=\frac{70\times \sin 45^{\circ }}{9.8}=\frac{70\times 0.707}{9.8}\approx 5\text{s}$$
• Horizontal distance: $$x=u\cos \theta \cdot t=70\times \cos 45^{\circ }\times 5=70\times 0.707\times 5=247.45\text{m}\approx 250\text{m}$$ (The slight discrepancy is due to rounding; exact calculation confirms $x=250\text{m}$.)

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Final Answer

• Velocity of projection: $70\text{m/s}$
• Direction of projection: $45^{\circ }$ to the horizontal