What is the linear velocity of a person at equator of the earth to its spinning motion ?

Q: What is the linear velocity of a person at equator of the earth to its spinning motion ?  (Radius of the earth = 6400 km).

Sol. The earth completes one rotation in 24 hour. Its angular velocity.

$\large \omega = \frac{2\pi N}{t}$

$\large \omega = \frac{2\pi \times 1}{24 \times 60 \times 60} =\frac{\pi}{43,200} rad/s$

The linear velocity,

$\large v = r \omega = 6.4 \times 10^6 \times \frac{\pi}{43,200}$

= 465.5 m/s

When a motor cyclist takes a U- turn in 4s what is the average angular velocity of the motor cyclist.

Q: When a motor cyclist takes a U- turn in 4s what is the average angular velocity of the motor cyclist.

Sol. When the motor cyclist takes a U – turn, angular displacement, θ = π rad and t = 4s.

The average angular velocity,

ω= θ/t = π /4=0.7855 rad /s

A projectile has the maximum range of 500 m. if the projectile is now thrown up on an inclined plane of 30° with the same speed…

Q: A projectile has the maximum range of 500 m. if the projectile is now thrown up on an inclined plane of 30° with the same speed, what is the distance covered by it along the inclined plane ?

Sol: $\large R_{max} = \frac{u^2}{g}$

$\large 500 = \frac{u^2}{g}$

$\large u^2 = 500 g$

$\large v^2 = u^2 + 2as$

$\large 0 = 500 g + 2(-gsin30^o)s$

s = 500 m

From the top of a tower, two balls are thrown horizontally with velocities u1 and u2 in opposite directions. If their velocities are perpendicular…

Q: From the top of a tower, two balls are thrown horizontally with velocities u1 and u2 in opposite directions. If their velocities are perpendicular to each other just before they strike the ground, find the height or tower.

Sol: Let h = height of the tower

$\large t = \sqrt{\frac{2 h}{g}}$

At time of reaching ground velocities are $\vec{v_1} = u_1 \hat{i} – g t\hat{j}$ and $\vec{v_2} = – u_2 \hat{i} – g t\hat{j}$

As velocities are perpendicular to each other hence ,

$\large \vec{v_1} . \vec{v_2} = 0$

$\large (u_1 \hat{i} – g t\hat{j}) . (- u_2 \hat{i} – g t\hat{j}) = 0$

$\large u_1 u_2 = g^2 t^2$

$\large t = \sqrt{\frac{u_1 u_2}{g^2}}$

$\large \sqrt{\frac{2 h}{g}} = \sqrt{\frac{u_1 u_2}{g^2}}$

$\large h = \frac{u_1 u_2}{g}$

A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. if the gun can impart a velocity of 500 m/sec to the bullet…

Q: A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. if the gun can impart a velocity of 500 m/sec to the bullet, at what height above the bird must he aim gun in order to hit it ?

Sol. x = vt or 100 = 500 × t ; t = 0.2 sec.

Now h = 0 + (1/2) × 10 × (0.2)2

= 0.20 m = 20 cm

A golfer standing on the ground hits a ball with a velocity 52 m/s at an angle θ Above the horizontal if tan⁡θ =  5/12 find…

Q: A golfer standing on the ground hits a ball with a velocity 52 m/s at an angle θ Above the horizontal if tan⁡θ =  5/12 find the time for which the ball is at least 15 m above the ground ? (g = 10 m/s2 )

Sol:  $\large u_y = u sin\theta$

$\large v_y = \sqrt{u_y^2 – 2 g y }$

$\large v_y = \sqrt{(u sin\theta)^2 – 2 g y }$

$\large v_y = \sqrt{(52 \times \frac{5}{13})^2 – 2 \times 10 \times 15 }$

vy = 10 m/s

$\large \Delta t = \frac{2 v_y}{10} = \frac{2 \times 10}{10}$

= 2 sec

In the absence of wind the range and maximum height of a projectile were R and H. If wind imparts of horizontal acceleration…

Q: In the absence of wind the range and maximum height of a projectile were R and H. If wind imparts of horizontal acceleration a = g/4 to the projectile then find the maximum range and maximum height.

Sol. H’ = H (since , u sin⁡θ remain same )

T’ = T

$\large R’ = u_x T + \frac{1}{2} a T^2$

$\large R’ = R + \frac{1}{2} (\frac{g}{4}) T^2$

$\large R’ = R + \frac{1}{2} g T^2$

R’  = R + H (since H’ =H)