SOLUTION: using the formula for a path of a projectile h = -4.9t^2 + vt + s where h is the height in meters, t is the time in seconds, v is the initial vertical velocity in meters per second
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Question 451571: using the formula for a path of a projectile h = -4.9t^2 + vt + s where h is the height in meters, t is the time in seconds, v is the initial vertical velocity in meters per second and s is the initial height in meters. A rocket fired from a height of 1.3 meters at an initial vertical velocity of 45 meters per second will the rocket ever reach a height of 190 meters?
I used h = -4.9t^2 + 45t + 1.3 and have come up with .09 and 9.21 but don't understand what that means. is that the max height or ?? Thanks
You can put this solution on YOUR website! h = -4.9t^2 + vt + s
Given: v = 45 m/s, s = 1.3 m
To find the maximum height, we take the derivative and set it equal to zero:
dh/dt = 0 = -9.8t + v
So t = v/9.8 = 45/9.8 = 4.592
Substituting the value for t into the original equation gives:
h = -4.9(4.592)^2 + 45(4.592) + 1.3
This gives h = 104.62 m
So the rocket never reaches a height of 190 m.
The graph of the trajectory is below:
