Answered: Physics Question

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Expert Solution

Step 1

If an object is thrown obliquely near the Earth's surface, then it moves in a curved path. The object is called as projectile and its motion is called as projectile motion. Acceleration of the projectile is almost constant and is equal to the acceleration due to gravity, g.

Step 2

NOTE- Consider downward motion to be negative

Here,
Vertical displacement covered by the projectile is $∆ y = - 210 m$
Horizontal displacement covered by projectile is $∆ x = 420 m$
Acceleration is, $a = - g = - 9.8 m / s^{2}$

Therefore, time taken to cover the vertical displacement is given by-
$\Rightarrow ∆ t = \sqrt{\frac{2 ∆ y}{a}} \Rightarrow ∆ t = \sqrt{\frac{2 \times 210}{9.8}} \Rightarrow ∆ t = 6.54 s$

Since time taken to fall vertically is the same time taken to cover the horizontal displacement.
$\Rightarrow ∆ t = \frac{∆ x}{v_{x}} \Rightarrow 6.54 s = \frac{420 m}{v_{x}} \Rightarrow v_{x} = 64.2 m / s$