Two projectiles are launched from the ground, and both reach the same vertical height. However, projectile B travels twice the horizontal distance as projectile A before hitting the ground.

a. How large is the vertical component of the initial velocity of projectile B compared with the vertical component of the initial velocity of projectile A?

b. How large is the horizontal component of the initial velocity of projectile B compared with the horizontal component of the initial velocity of projectile A?

c. Suppose projectile A is launched at an angle of 45 degrees to the horizontal. What is the ratio, VB/VA, of the speed of projectile B, VB, compared with the speed of projectile A, VA?

If they travel the same vertical height, they had to have had the same initial vertical velocity

If one traveled twice the horizontal distance in the same time as the other, it had to have twice the horizontal velocity.

Of Va= sqrt (Vv^2 + Vh^2, and Vb=sqrt( Vv^2 + (2Vh)^2) (see above), but at 45 deg, Vv^2=Vh^2,

when you take the ratio, it reduces. I will be happy to check your work.

1:2

do you still need the computation for this one?.

I need to get the answer instantly

lol

a. The vertical component of the initial velocity of projectile B is the same as the vertical component of the initial velocity of projectile A. In other words, they are equal.

b. The horizontal component of the initial velocity of projectile B is twice the horizontal component of the initial velocity of projectile A. So, the horizontal component of the initial velocity of projectile B is twice as large as that of projectile A.

c. If projectile A is launched at an angle of 45 degrees, the vertical component of its initial velocity is equal to the horizontal component of its initial velocity. Since we established that projectile B has twice the horizontal component of projectile A, then projectile B has twice the speed of projectile A.

So, the ratio, VB/VA, of the speed of projectile B compared with the speed of projectile A is 2:1. Projectile B is twice as fast as projectile A.

a. The vertical component of the initial velocity of projectile B is the same as the vertical component of the initial velocity of projectile A. This is because both projectiles reach the same vertical height, indicating that they both have the same upward vertical velocity component at the start.

b. The horizontal component of the initial velocity of projectile B is twice the horizontal component of the initial velocity of projectile A. This is because projectile B travels twice the horizontal distance compared to projectile A before hitting the ground. Therefore, the initial horizontal velocity of projectile B must be twice the initial horizontal velocity of projectile A.

c. Suppose projectile A is launched at an angle of 45 degrees to the horizontal. In this case, the vertical component of the initial velocity (Vv) is equal to the horizontal component of the initial velocity (Vh) since Vv^2 = Vh^2.

Using this information, we can find the ratio of the speed of projectile B (VB) to the speed of projectile A (VA).
VB = sqrt(Vv^2 + (2 * Vh)^2) = sqrt(Vh^2 + (2 * Vh)^2) = sqrt(5 * Vh^2) = sqrt(5) * Vh
VA = sqrt(Vv^2 + Vh^2) = sqrt(Vh^2 + Vh^2) = sqrt(2 * Vh^2) = sqrt(2) * Vh

Therefore, the ratio VB/VA is sqrt(5) * Vh / (sqrt(2) * Vh) = sqrt(5/2).

So, the ratio of the speed of projectile B to the speed of projectile A is sqrt(5/2) or approximately 1.58.