Home
Class 11
PHYSICS
The equation of projectile is y=16x-(x^(...

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

A

16 m

B

8 m

C

64 m

D

12.8 m

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

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

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

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

The equation of a projectile is y = sqrt(3)x - ((gx^2)/2) 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 a trajectory of a projectile is y=8x-4x^(2) The maximum height of the projectile is (A) 8m (B) 6m (C) 2m (D) 4m

The equation of the path of the projectiles is y=0.5x-0.04x^(2) .The initial speed of the projectile is? (g=10ms^(-2))

[" The equation of trajectory of a projectile is "],[y=ax-bx^(2)" where "x" and "y" axes are along "],[" horizontal and vertical directions "],[" respectively.The range of the projectile is "]

The equation of trajectory of a projectile is y=10x-(5/9)x^2 if we assume g=10ms^(-3) then range of projectile (in metre) is

The equation of a trajectory of a projectile is y=8x-4x^(2) . The maximum height of the projectile is ?