Lesson_1.4_Homework_Solution

# Lesson_1.4_Homework_Solution - Lesson 1.4 Homework...

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Lesson 1.4 Homework Solutions 1. A pitcher throws a fast ball with a horizontal velocity of 125 kmh -1 towards the home plate which is 18.5 m away. Neglecting air resistance, find (i) the time taken by the ball to reach just above the home plate, (ii) the vertical drop of the ball during this time. (i) The ball has a horizontal velocity of 125 km.h -1 = 34.7 m.s -1 . This horizontal velocity remains uniform and it travels a horizontal distance of 18.5 with this uniform velocity. The time taken to do this = distance/uniform velocity = 18.5/34.7 = 0.53 s (ii) During this time of 0.53 seconds the ball is falling freely with an initial vertical velocity zero and an acceleration of –9.8 m/s 2 . The vertical fall is given by y = (1/2)gt 2 = (1/2) × 9.8 × 0.53 2 = 1.38 m 2. A spring loaded gun placed on a table 1.0 m above the ground level projects a marble with an initial speed of 4.8 ms -1 . (i) What is the time taken by the marble to drop to the ground? (ii) How far from the table will the ball fall? ( i) Consider the vertical motion: The initial position y o =1.0 m, when it falls to the ground, the final position y = 0, initial vertical velocity v o = 0 y = y o + vot + ½ at 2 0 = 1.0 m + 0 + ½(-9.8)t 2 = 1 – 4.9 t 2 4.9t 2 = 1 t 2 = 0.2 s 2 t = 0.45 s (ii) During this time t = 0.45 s, the ball is moving horizontally with a uniform velocity of 4.8 ms -1 . The horizontal displacement is: x = vt = 4.8 ms -1 × 0.45 s = 2.2 m 3. A ball rolls with a constant velocity of 1.5 ms -1 at an angle of 45 o relative to the x-axis in the 4 th quadrant. Taking the ball to be at the origin at t = 0, what is its position ( x and y co-ordinates) at t = 1.65 s? What is the displacement of the ball at this time?

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45 o 1.5 m /s The figure above shows the velocity of the ball in the 4 th quadrant at an angle 45 o with the x-axis. This angle is negative since it is measured clockwise with the x-axis. Therefore we have a velocity which has a magnitude 1.5 m/s and an angle –45 o . The x and y components of this velocity is given by: v x = v Cos θ = 1.5 Cos(-45) = 1.1 m.s -1 v y = v Sin θ = 1.5 Sin (-45) = -1.1 m.s -1 Now we use the displacement equations in the x and y directions x = x o + v xo t + (1/2)a t 2 Since the ball at the origin at t = 0, x o = 0. The velocity is a constant, therefore, v xo = 1.1 m/s and a = 0. Also given is t = 1.65 s. Using these values in the equation gives x = 0 + (1.1 m/s)(1.65 s) = 1.8 m Similarly, y = y o + v yo t + (1/2)a t 2 where v yo = -1.1 m.s -1 Therefore, y = 0 - (1.1)(1.65) = -1.8 m At t= 1.65 s, the ball is at x = 1.8m, y = -1.8 m, at the point (1.8, -1.8).
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## This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.

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Lesson_1.4_Homework_Solution - Lesson 1.4 Homework...

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