A stone is thrown with an initial speed of 15 m/s at an angle of 53 $\degree$ above the horizontal from the top of a 35 m building.If g = 9.8 m/s² and air resistance is negligible,what is the magnitude of the horizontal displacement of the rock? A)38 m B)46 m C)66 m D)90 m

Question

Quizplus · Accepted Answer

The horizontal displacement of the rock is given by:$$d = v_0 \cos(	heta) t$$where:- $v_0$ is the initial speed of the rock- $	heta$ is the angle of the rock's trajectory above the horizontal- $t$ is the time of flight of the rockThe time of flight of the rock can be found by using the following equation:$$t = \frac{2v_0 \sin(	heta)}{g}$$where:- $g$ is the acceleration due to gravitySubstituting the given values into the above equations, we get:$$d = (15 \ m/s) \cos(53^\circ) \left(\frac{2(15 \ m/s) \sin(53^\circ)}{9.8 \ m/s^2}ight) = 38 \ m$$Therefore, the magnitude of the horizontal displacement of the rock is 38 m.

A Stone Is Thrown with an Initial Speed of 15 °\degree°

A Stone Is Thrown with an Initial Speed of 15 $\degree$