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1. A baseball is seen to pass upward by a window 28 m above the street with a ve

ID: 2003865 • Letter: 1

Question

1. A baseball is seen to pass upward by a window 28 m above the street with a vertical speed of 10 m/s. The ball was thrown from the street.

(a) What was its initial speed?
m/s
(b) What altitude does it reach?
m
(c) How long after it was thrown did it pass the window?
s
(d) After how many more seconds does it reach the street again?
s

2. You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -60.0 units. What are the two possibilities for its x component?


(a) Give the resultant in terms of components.
Rx =  
Ry =  

(b) What is the magnitude of the resultant?

What is the resultant's angle above the +x axis?
°

Explanation / Answer

Ans.-1   (a) What was its initial speed?

Let, initial speed of the ball = u

As per equation of motion: v2= u2 + 2ah

Here u = initial velocity

a = -g = acceleration due to gravity (it is acting in the opposite direction of motion)

h = 28 m (Height of window)

v=10 m/s at window

putting the value in the equation: v2=u2+2* (-g)*h

u2 = v2 - 2gh

Here g= 9.8 m/s2

u2= 102 - (2*9.8*28)

u = sqrt (648.8)

u= 25.47 m/s

(b) What altitude does it reach?

As per equation of motion: v2= u2 + 2ah

v = 0 at the peak point

u = 25.47 m/s

a = -g = -9.8 m/s2

putting the values in the above equation: 0 = (25.47)2 – 2* (9.8)* h

h = (25.47)2/ (2*9.8)

h = 33.1 m

(c) How long after it was thrown did it pass the window?

As per equation of motion: v = u + at

here v= 10 m/s

u= 25.47 m/s

a =-g =-9.8 m/s2

putting the value in the equation: 10 = 25.47- 9.8* t

t = (25.47 – 10)/ 9.8

t = 1.58 s

(d) After how many more seconds does it reach the street again?

As per the equation of motion: S = ut + ½*at2

S= displacement = 0

Putting the values in the equation-

0 = 25.47t -0.5* 9.8t2

4.9t2-25.47t = 0                                                       

t=0 &

t =25.47/4.9

t= 5.2 s

Value of displacement is ZERO, 2 times in this motion.

Ans 2- Let, vector = v

Magnitude of vector |v|= sqrt (x2 + y2)

|y|= -60

|x|2 = |v|2- |y|2

|x|2 = 902 – (-60)2

|x|= sqrt (8100-3600)

|x| = 67.1

Ans 3 - (a) Give the resultant in terms of components.

Rx = algebraic sum of components of all the vectors on x-axis

Rx = Ax + Bx + Cx

Rx = 60 cos 28 + (-40 cos 50) + 0                   (Bx is along –ve x-axis)

Rx = 52.98 – 25.71

Rx = 27.27

Ry = algebraic sum of components of all the vectors on y-axis

Ry = Ay + By + Cy

Ry = 60 sin 28 + (40 sin 50) + (- 46.8)                   (Cy is along –ve y-axis)

Ry = 28.17 + 30.64 – 46.8

Ry = 12.01

(b) What is the magnitude of the resultant?

R = sqrt (Rx2+ Ry2)

R = sqrt {(27.27)2 + (12.01)2}

R = sqrt (743.6529 + 144.2401)

R = sqrt (887.893)

R = 29.78

(c) What is the resultant's angle above the +x axis?

Angle x ° = tan-1 (Ry / Rx )

x ° = tan-1 (12.01/ 27.27)

x ° = tan-1 (0.443)

x ° = 23.89°