Question

1. a projectile is launched horizontally at a speed of 30 meters per second from a platform located a vertical distance h above the ground. the projectile strikes the ground after time t at horizontal distance d from the base of the platform. neglect friction. v = 30. m/s impact location d if total time of flight is 2.5s, what is the height of the cliff?

Explanation:

Step1: Identify vertical motion formula

For free fall (initial vertical velocity $v_{0y}=0$), the vertical displacement formula is $h = v_{0y}t + \frac{1}{2}gt^2$.

Step2: Substitute known values

$v_{0y}=0$, $g=9.8\ \text{m/s}^2$, $t=2.5\ \text{s}$.
$h = 0 + \frac{1}{2} \times 9.8 \times (2.5)^2$

Step3: Calculate the height

First compute $(2.5)^2=6.25$, then $\frac{1}{2} \times 9.8 = 4.9$, finally $4.9 \times 6.25 = 30.625$

Answer:

$30.625\ \text{meters}$ (or rounded to $31\ \text{meters}$ for practical purposes)