NESA Physics Projectile Motion

This is the initial vertical velocity, calculated using $v_y = v \sin(\theta) = 38 \sin(42^\circ) \approx 25 \text{ m s}^{-1}$.

The horizontal component of velocity is found using the cosine of the angle: $v_x = v \cos(\theta) = 38 \cos(42^\circ) \approx 28.2 \text{ m s}^{-1}$.

This value is incorrect; it does not result from applying the trigonometric component formulas to the given velocity and angle.

This is impossible because a component of the velocity vector ($v_x$) cannot be greater than the total magnitude of the velocity ($38 \text{ m s}^{-1}$).

This is the correct maximum height derived from the formula $H = \frac{v^2 \sin^2(\theta)}{2g}$. Substituting the values yields $\frac{(15)^2 (0.5)^2}{2(9.8)} \approx 2.87 \text{ m}$.

This option is incorrect; it may result from a calculation error or using an incorrect fraction for the vertical component.

This answer incorrectly uses the horizontal component (cosine) instead of the vertical component (sine), calculating $\frac{v^2 \cos^2(\theta)}{2g}$.

This option ignores the launch angle and calculates the height as if the projectile were fired straight up ($90^\circ$) using $\frac{v^2}{2g}$.

Correct. The time of flight is determined by the vertical drop using $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 46}{9.8}} \approx 3.06$ s. The horizontal displacement is then calculated as $d = v_x \times t = 25 \times 3.06 \approx 76.6$ m, which rounds to 77 m.

Incorrect. This value likely results from incorrectly calculating the time of flight using the linear ratio $t = \frac{h}{g}$ instead of the kinematic square root formula, yielding $t \approx 4.7$ s and a distance of roughly 117 m.

Incorrect. This large displacement implies a time of flight of roughly 7.6 s. This would be the correct answer if the projectile were on the Moon ($g \approx 1.6$ m s$^{-2}$), but it is incorrect for Earth gravity.

Incorrect. This answer is close to the result obtained if one forgets to take the square root when solving for time ($t = \frac{2h}{g} \approx 9.4$ s), which leads to a much larger horizontal displacement.