15. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s–1 can go without hitting the ceiling of the hall?
Solution:
Speed of the ball, u = 40 m/s
Maximum height, h = 25 m
In projectile motion, the maximum height reached by a body projected at an angle θ, is given by the relation:
h = \(\frac {u^2sin^2\theta}{2g} \)
25 = \(\frac {(40)^2 sin^2\theta}{2 \times 9.8}\)
sin2 θ = 0.30625
sin θ = 0.5534
θ = sin–1
(0.5534) = 33.60°
Horizontal range, R = \(\frac {u^2sin2\theta}{g} \)
= \(\frac {(40)^2 \times sin \, 2 \times 33.60}{9.8}
\)
= \(\frac {1600 \times sin\,67.2}{9.8}
\) = \(\frac {1600 \times 0.922}{9.8}
\) = 150.53 m
16. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Solution:
Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R = \(\frac {u^2 sin\, 2\theta}{g}\)
100 = \(\frac {u^2 }{g}\) sin 90°
\(\frac {u^2 }{g}\) = 100 .....(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H. Acceleration, a = –g
Using the third equation of motion:
v2 - u2 = -2gH
H = \(\frac 12 \times \frac {u^2}{g} = \frac 12\) x 100 = 50 m
17. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Solution:
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency, v = \(\frac {Number\,of \,revolutions}{Time taken} = \frac {14}{25}Hz
\)
Angular frequency, ω = 2πν = \(2 \times {22}{7} \times {14}{25} = \frac {88}{25} rad \,s^{-1}\)
Centripetal acceleration, ac = ω2 r = \((\frac {88}{25})^2 \times 0.8 = 9.91 \,m/s^2\)
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
18. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:
Radius of the loop, r = 1 km = 1000 m
Speed of the aircraft, v = 900 km/h = 900 x \(\frac {5}{18}\) = 250 m/s
Centripetal acceleration, ac = \(\frac {v^2}{r}\) = \(\frac {(250)^2}{1000}\) = 62.5 m/s2
Acceleration due to gravity, g = 9.8 m/s2
\(\frac {a_c}{g} = \frac {62.5}{9.8}\) = 6.38
ac = 6.38 g
19. Read each statement below carefully and state, with reasons, if it is true or false:
a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Solution:
a) False The net acceleration of a particle in circular motion is not always directed along the radius of the circle toward the centre. It happens only in the case of uniform circular motion.
b) True At a point on a circular path, a particle appears to move tangentially to the circular path. Hence, the velocity vector of the particle is always along the tangent at a point.
c) True In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.
20. The position of a particle is given by
\(\overrightarrow r = 3.0t \,\hat i - 2.0t^2 \hat j + 4.0 \hat k \,m\)
Where t is in seconds and the coefficients have the proper units for r to be in meters.
a) Find the v and a of the particle?
b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?
Solution:
a) The position of the particle is given by:
\(\overrightarrow r = 3.0t \,\hat i - 2.0t^2 \hat j + 4.0 \hat k \,m\)
Velocity \(\overrightarrow v\), of the particle is given as:
∴ \(\overrightarrow v\) = \(\frac {d\overrightarrow r}{dt} = \frac {d}{dt} (3.0t\, \hat i - 2.0t^2 \hat j + 4.0 \,\hat k)\)
∴ \(\overrightarrow v\) = \(3.0 \hat i - 4.0t \, \hat j\)
Acceleration \(\overrightarrow a\), of the particle is given as:
\(\overrightarrow a\) = \(\frac {d\overrightarrow v}{dt} = \frac {d}{dt} (3.0t\, \hat i - 4.0 \,\hat j)\)
\(\overrightarrow a\) = \(-4.0\, \hat j\)
8.54 m/s, 69.45° below the x-axis
b) We have velocity vector, \(\overrightarrow v\) = \(3.0 \hat i - 4.0t \, \hat j\)
At t = 2.0 s:
\(\overrightarrow v\) = \(3.0 \hat i - 8.0 \, \hat j\)
The magnitude of velocity is given by:
|\(\overrightarrow v\)| = \(\sqrt {3^2 + (-8)^2} = \sqrt{73} \) = 8.54 m/s
Direction, θ = tan-1 \((\frac {v_y}{v_x})\)
= tan-1\((\frac {-8}{3})\) = tan-1(2.667)
= -69.45°
The negative sign indicates that the direction of velocity is below the x-axis.