Question

the equation of trajectory of a projectile is given by y= √3x-5x^2.[g=10ms^-1]
1) then the angle of projection?
2) initial velocity of the projectile?
3) time of ascent of projectile?





pls answer my question with correct answer ASAP..​

Answer 1

Answer:

1) y=xtanθ–

2v

0

2

cos

2

θ

gx

2

So, by comparing the given equation

y=

3

x−5x

2

We get that tanθ=

3

θ=60

°

So, Angel of projection is 60°

2) The equation of trajectory of a projectile is

y=10x−

9

5

x

2

For the range of projectile,

x=R and y=0

0=10R−

9

5

R

2

R(10−

9

5

R)=0

10−

9

5

R=0

5R=10×9

R=18m

3) Sorry I don't know that