Question

use the image to answer the question. a stunt motorcyclist jumps off a horizontal platform with an initial velocity of 30 m/s. if the height of the platform is 2 m high, how far from the base of the platform will the motorcyclist land? (hint: find δt first.) (1 point) no answer. it’s not possible to take the square root of a negative number. 19.2 m 0.64 m 0.12 m

Explanation:

Step1: Find time - fall in y - direction

The vertical - motion is a free - fall motion. The initial vertical velocity $v_{0y}=0\ m/s$, and the vertical displacement $\Delta y = 2\ m$, and the acceleration due to gravity $g = 9.8\ m/s^{2}$. Using the equation $\Delta y=v_{0y}t+\frac{1}{2}gt^{2}$, since $v_{0y} = 0\ m/s$, the equation simplifies to $\Delta y=\frac{1}{2}gt^{2}$. Then $t=\sqrt{\frac{2\Delta y}{g}}$.
$t=\sqrt{\frac{2\times2}{9.8}}\ s\approx\sqrt{\frac{4}{9.8}}\ s\approx0.64\ s$

Step2: Find horizontal distance

The horizontal motion is a uniform - motion with constant velocity because there is no horizontal acceleration ($a_x = 0$). The horizontal velocity $v_{0x}=30\ m/s$. Using the equation $x = v_{0x}t$.
$x=30\times0.64\ m = 19.2\ m$

Answer:

19.2 m