A shell is launched from level ground into the air. It reaches a maximum height of 45\text… — JAMB UTME

A shell is launched from level ground into the air. It reaches a maximum height of $45\text{ m}$ above the ground and has a total horizontal range of $240\text{ m}$ . Taking $g = 10\text{ m/s}^2$ , calculate the magnitude of the initial launch velocity of the shell in $\text{m/s}$ .

Answer: 50 m/s

Answer

The initial launch velocity of the shell is

50\text{ m/s}

Combining the expressions for maximum height

H = \frac{u^2\sin^2\theta}{2g}

and range

R = \frac{2u^2\sin\theta\cos\theta}{g}

gives

\tan\theta = \frac{4H}{R}

. With

H = 45\text{ m}

and

R = 240\text{ m}

, we get

\tan\theta = 0.75 = \frac{3}{4}

, which yields

\sin\theta = 0.6

. Substituting these into the height equation yields

45 = \frac{u^2(0.6)^2}{20}

, solving to

u = 50\text{ m/s}

Step-by-Step Solution

Express the launch angle in terms of maximum height and horizontal range

\tan\theta = \frac{4H}{R} = \frac{4 \times 45}{240} = 0.75

Dividing the maximum height formula

H = \frac{u^2\sin^2\theta}{2g}

by the horizontal range formula

R = \frac{2u^2\sin\theta\cos\theta}{g}

yields

\frac{H}{R} = \frac{1}{4}\tan\theta

Find the sine of the launch angle from the tangent value

\sin\theta = 0.6

For a right-angled triangle with

\tan\theta = \frac{3}{4}

, the hypotenuse is

\sqrt{3^2 + 4^2} = 5

, giving

\sin\theta = \frac{3}{5} = 0.6

Calculate the magnitude of the initial velocity

u

u = 50\text{ m/s}

Substituting values into

H = \frac{u^2\sin^2\theta}{2g}

gives

45 = \frac{u^2 (0.6)^2}{2(10)} \implies 900 = 0.36 u^2 \implies u = 50\text{ m/s}

Key Concept

Interdependence of Maximum Height, Range, and Launch Velocity in Projectile Motion

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic