Home
Class 11
PHYSICS
Equation of a projectile is given by y=P...

Equation of a projectile is given by y=`Px-Qx^(2)` where P,Q are constants, the ratio of maximum height to Range of the projectile is

A

`(P^(2))/(Q)`

B

4P

C

`(P)/(4)`

D

`(Q^(2))/(2P)`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

Equation of a projectile is given by y = Px - Qx^(2) , where P and Q are constants. The ratio of maximum height to the range of the projectile is

The equation of a projectile is y = ax - bx^(2) . Its horizontal range is

The equation of projectile is y=16x-(x^(2))/(4) the horizontal range is:-

The equation of motion of a projectile is y = 12 x - (3)/(4) x^2 . The horizontal component of velocity is 3 ms^-1 . What is the range of the projectile ?

The equation of motion of a projectile is y = 12 x - (3)/(4) x^2 . The horizontal component of velocity is 3 ms^-1 . What is the range of the projectile ?

The equation of a projectile is y = sqrt(3)x - ((gx^2)/2) the horizontal range is

A projectile is projected from the ground by making an angle of 60^@ with the horizontal. After 1 s projectile makes an angle of 30^@ with the horizontal . The maximum height attained by the projectile is (Take g=10 ms^-2)

The equation of projectile is y = 16 x - (5 x^2)/(4) . Find the horizontal range.

A projectile has a range of 40 m and reaches a maximum height of 10 m . Find the angle at which the projectile is fired.

A projectile has a range of 40 m and reaches a maximum height of 10 m. Find the angle at which the projectile is fired.