Equation of trajector of ground to ground projectile is `y=2x-9x^(2)`. Then the angle of projection with horizontal and speed of projection is : `(g=10m//s^(2))`
A
`theta_(0)=sin^(-1)((1)/(sqrt(5)))` and `v_(0)=(5)/(3)ms^(-1)`
B
`theta_(0)=cos^(-1)((2)/(sqrt(5)))` and `v_(0)=(3)/(5)ms^(-1)`
C
`theta_(0)=cos^(-1)((1)/(sqrt(5)))` and `v_(0)=(5)/(3)ms^(-1)`
D
`theta_(0)=sin^(-1)((2)/(sqrt(5)))` and `v_(0)=(3)/(5)ms^(-1)`
The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is
The equation of a projectile is y=sqrt(3)x-(gx^(2))/(2) the angle of projection is:-
The equations of motion of a projectile are given by x=36tm and 2y=96t-9.8t^(2)m .The angle of projection is
The equations of motion of a projectile are given by x=36tm and 2y =96t-9.8t^(2)m . The angle of projection is
A stone projected from ground level falls on the ground after "4" seconds.Then the height of the stone 1 second after the projection is (g=10m/s^(2))
The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Angle of projection theta is :-
A projectile is thrown with velocity u making an angle theta with the horizontal. Its time of flight on the horizontal ground is 4 second. The projectile is moving at angle of 45^(@) with the horizontal just one second after the projection. Hence the angle theta is (take g=10m//s^(2) )
For ground to ground projectile motion equation of path is y=12x-3//4x^(2) . Given that g=10ms^(-2) . What is the range of the projectile ?
The parabolic path of a projectile is represented by y=x/sqrt3-x^(2)/60 in MKS units: Its angle of projection is (g=10ms^(-2))
