A projectile is launched from ground level with an inital velocity of 50 m/s at an angel of 60 degreess aove the horizontal. Use g=10m/s. What is the max height reached by the projectile?

Question

A projectile is launched from ground level with an inital velocity of 50 m/s at an angel of 60 degreess aove the horizontal. Use g=10m/s. What is the max height reached by the projectile?

I used change in y=voyt +1/2at^2
=50(2.4) + 1/2(=10) (2.5)^2
=125-31.25
=93.75m

I was just wondering if i did this correctly.

b) What is the vertical component of the final velocity right before the projectile returns to the ground?

I used vy= v sin theda
but my roblem with this is what would v equal to ?

Answer 1

a) yes, but I wonder how you got the time to max height.

b) I have no idea what "right before...returns to ground" means. The wording is very poor. In physics, we consider gravity returning the projectile to ground starting at launch and working until it strikes the ground.

Answer 2

For the first part of your question, finding the maximum height reached by the projectile, your calculation is correct.

To solve part b), we need to find the vertical component of the final velocity right before the projectile returns to the ground.

To start, let's break down the initial velocity into its horizontal and vertical components. The initial velocity is given as 50 m/s at an angle of 60 degrees above the horizontal. We can find the vertical component of the initial velocity (V_y initial) using the equation V_y initial = V initial * sin(theta), where V initial is the magnitude of the initial velocity and theta is the launch angle.

So, V_y initial = 50 m/s * sin(60 degrees).
V_y initial = 50 m/s * (sqrt(3)/2).
V_y initial = 25(sqrt(3)) m/s.

Now, as the projectile returns to the ground, the vertical component of the final velocity (V_y final) would be equal in magnitude but opposite in direction to the initial vertical velocity. Therefore, V_y final = -25(sqrt(3)) m/s.

Note that the horizontal component of the final velocity (V_x final) would be the same as the initial horizontal velocity since there is no horizontal force acting on the projectile. So, V_x final = V_x initial = V initial * cos(theta).

Therefore, the vertical component of the final velocity right before the projectile returns to the ground is -25(sqrt(3)) m/s.

Remember, when solving projectile motion problems, it's crucial to break down the initial velocity into its components (horizontal and vertical) and then consider the effects of gravity on each component separately.