During a school funfair, a student throws a wet sponge at a teacher - Scottish Highers Physics - Question 1 - 2018

The vertical component of the initial velocity can be calculated using the formula:

$u_{y} = u imes ext{sin}( heta)$

Where:

• $u = 7.4 \, m/s$
• $\theta = 30°$

Thus:

$u_{y} = 7.4 \times \text{sin}(30°) = 3.7 \, m/s$

The time to reach maximum height can be determined using the formula:

$t = \frac{u_{y}}{g}$

Where:

• $u_{y} = 3.7 \, m/s$ (initial vertical velocity)
• $g = 9.8 \, m/s^{2}$ (acceleration due to gravity)

Thus:

$t = \frac{3.7}{9.8} \approx 0.38 \, s$

The total time for the sponge to hit the teacher includes both the ascent and descent:

Total time, $T = t + 0.45 = 0.38 + 0.45 = 0.83 \, s$

Next, we calculate the maximum height reached:

$h_{max} = 1.5 + (u_{y} \times t) - \frac{1}{2} g t^2$

Substituting the values:

$h_{max} = 1.5 + (3.7 \times 0.38) - \frac{1}{2} \times 9.8 \times (0.38)^2$

Calculating: $h_{max} \approx 1.5 + 1.406 - 0.707 = 2.199 \, m$

Finally, for the height at which the sponge hits the teacher:

$h = h_{max} - (0.5 \times g \times (0.45)^2)$

Thus:

determining height: $h \approx 2.199 - (0.5 \times 9.8 \times 0.2025) \approx 2.199 - 0.99 \approx 1.21 \, m$

The student's statement is incorrect because, while an increase in speed may suggest a decrease in time taken, this is not the case if the angle remains unchanged. A higher initial speed increases both the horizontal and vertical displacement. The sponge will reach a higher altitude and travel a greater horizontal distance, which can result in the same or more time to reach the same horizontal position.