Question

A projectile with speed \(50\ \text{m s}^{-1}\) is thrown at an angle of \(60^\circ\) with the horizontal. The maximum height that can be reached is
\[ \text{Acceleration due to gravity}=10\ \text{m s}^{-2} \]

• A. \(90.75\ \text{m}\)
• B. \(70.00\ \text{m}\)
• C. \(85.00\ \text{m}\)
• D. \(93.75\ \text{m}\)

Hint:

For maximum height of a projectile, use only the vertical component of velocity: \(H=\frac{u^2\sin^2\theta}{2g}\).

Answer 1

The Correct Option is D

Step 1: Use the formula for maximum height.
For a projectile, maximum height is given by
\[ H=\frac{u^2\sin^2\theta}{2g} \]
Here,
\[ u=50\ \text{m s}^{-1},\qquad \theta=60^\circ,\qquad g=10\ \text{m s}^{-2} \]

Step 2: Substitute the values.
\[ H=\frac{50^2\sin^260^\circ}{2(10)} \]
Since
\[ \sin60^\circ=\frac{\sqrt3}{2} \]
\[ \sin^260^\circ=\frac34 \]
Therefore,
\[ H=\frac{2500\times \frac34}{20} \]
\[ H=\frac{1875}{20} \]
\[ H=93.75\ \text{m} \]

Step 3: Final conclusion.
Hence, the maximum height reached by the projectile is
\[ \boxed{93.75\ \text{m}} \]