When a body vertically projected with 200m/s. then velocity of body at490m
Wondering if the vertical velocity is only a Launch velocity...
final KE+finalPE=initial KE
1/2 m vf^2 +mg(490)=1/2 m 200^2
solve for vf
VF=sqrt(200^2-2*9.8*490)
To calculate the velocity of a vertically projected body at a specific height, we can use the equations of motion. In this case, we have the initial velocity (200 m/s) and the final position or height (490 m).
First, let's identify the known values:
- Initial velocity (u) = 200 m/s
- Final position (s) = 490 m
Next, we need to find the final velocity (v) at the given height. We can use the equation of motion that relates initial velocity, final velocity, acceleration, and displacement:
v^2 = u^2 + 2as
Rearranging the equation to solve for the final velocity (v):
v = sqrt(u^2 + 2as)
Substituting the known values:
v = sqrt((200 m/s)^2 + 2 * 9.8 m/s^2 * 490 m)
Simplifying:
v = sqrt(40000 m^2/s^2 + 9800 m^2/s^2 * 490 m)
v = sqrt(40000 m^2/s^2 + 4802000 m^2/s^2)
v = sqrt(4842000 m^2/s^2)
Calculating the final velocity:
v ≈ 2200 m/s
Therefore, when a body is vertically projected with an initial velocity of 200 m/s and reaches a height of 490 m, the velocity of the body at that height is approximately 2200 m/s.