A man is at height of 100 m. he sees a car which makes an angle of with man's position. If the car moves to a point where angle is what is the distance moved by it?
If ar and at represent radial and tangential accelerations, the motion of a particle will be uniformly circular if :
ar = 0 and at = 0
ar = 0 and at ≠ 0
ar ≠ 0 and at = 0
ar ≠ 0 and at ≠ 0
ar ≠ 0 and at = 0
If ar = 0 and at = 0, then motion is uniform translatory.
If ar = 0 but at ≠ 0, then motion is accelerated translatory.
If ar ≠ 0 but at = 0, then motion is uniform circular.
If ar ≠ 0 and at ≠ 0, then motion is a non-uniform circular.
An arrow is shot, its range is 200 m and its time of flight is 5 s. If g =10 m/s2, then the horizontal component of velocity of the arrow is :
Flight time =
u cosθ = 40 m/s (Horizontal velocity)
If vectors and are perpendicular to each other, then the positive value of a is
If two vectors are perpendicular to each other, then
⇒ a2 2a 3 = 0
Solving this quadratic equation, we get a = 3 or 1.
Therefore positive value of a is 3.
Which of the following velocity-time graphs shows a realistic situation for a body in motion?
Other graph shows more than one velocity of the particle at single instant of time which is not practically possible.
At the uppermost point of a projectile, its velocity and acceleration are at an angle of
At the uppermost point of a projectile, the vertical component of the velocity of projection becomes zero, while the horizontal component remains constant. And the acceleration (due to gravity) is always vertically downwards. Therefore at the uppermost point of a projectile, its velocity and acceleration are at an angle of 90°.
For angles of projection of a projectile at angles (45° - θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of
For complementary angles of projection, their horizontal ranges will be same.
We know that, horizontal ranges for complementary angles of projection will be same.
The projectiles are projected at angles and which are complementary to each other i.e., two angles add up to give . Hence, horizontal ranges will be equal. Thus, the required ratio is 1:1.
The maximum range of a gun on horizontal terrain is 16 km if g=10m/s2. What must be the muzzle velocity of the shell?
A lead shot of 1 mm diameter falls through a long column of glycerine. The variation of its velocity 'v' with distance covered is represented by
Initially due to the action of gravity, the lead shot will move with increasing velocity for some time. Then due to the viscosity of the glycerine column, the lead shot will attain a constant terminal velocity. As initially. there is some upthrust on the shot due to glycerine the increase of velocity will not be fully linear. So the variation is shown by plot (a).