# Answer to Question #2652 in Mechanics | Relativity for tom

Question #2652

A projectile is fired from the ground with an initial velocity of 82 m/s at an angle of 55 above the horizontial .

1- What are the initial x and y componets of the velocity ?

2- What is the x component of the projectile 1.2 s after being fired , relative the firing point?

3- What is the y component of the projectile 1.2 s after being fired , relative the firing point ?

1- What are the initial x and y componets of the velocity ?

2- What is the x component of the projectile 1.2 s after being fired , relative the firing point?

3- What is the y component of the projectile 1.2 s after being fired , relative the firing point ?

Expert's answer

1) Let Vx and Vy be the component s of the velocity.

Then their initial values are the following:

Vx(0)= 82*cos55 =47.03

Vy(0)= 82*sin 55=67.17

2) The move of projectile consists of two moves: along x-axis and along y-axis.

On the projectile act a gravitation force along y-axis.

Therefore along x-axis projectile moves with constant velocity& Vx(0), whence the x component of projectile after 1.2& s is equal to

Sx = Vx * t = 82*cos55 * 1.2 = 47.03*1.2 = 56.44

3) The move of projectile along y-axis has constant acceleration g = -9.8 m/s^2, since it is directed down, while Vy is directed up.

Hence y component of projectile after 1.2& s is equal to

Sx = Vx * t – g t^2 /2 =

& 82*sin55 * 1.2 – 9.8 (1.2)^2 / 2

= 67.17 *1.2 – 7.056 = 73.548

Then their initial values are the following:

Vx(0)= 82*cos55 =47.03

Vy(0)= 82*sin 55=67.17

2) The move of projectile consists of two moves: along x-axis and along y-axis.

On the projectile act a gravitation force along y-axis.

Therefore along x-axis projectile moves with constant velocity& Vx(0), whence the x component of projectile after 1.2& s is equal to

Sx = Vx * t = 82*cos55 * 1.2 = 47.03*1.2 = 56.44

3) The move of projectile along y-axis has constant acceleration g = -9.8 m/s^2, since it is directed down, while Vy is directed up.

Hence y component of projectile after 1.2& s is equal to

Sx = Vx * t – g t^2 /2 =

& 82*sin55 * 1.2 – 9.8 (1.2)^2 / 2

= 67.17 *1.2 – 7.056 = 73.548

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment