a model rocket is projected straight upward from the ground level. It is fired with an initial velocity of 192 ft/s. How high is the rocket after 10 seconds?
h = Vo*t + 0.5g*t^2
Vo = 192Ft/s
t = 10 s.
g = -32Ft/s^2
Solve for h.
To determine the height of the rocket after 10 seconds, we need to use the principles of vertical motion and the basic equations of motion.
First, we need to understand the acceleration acting on the rocket. Since the rocket is projected straight upward, the only force acting on it is gravity. We can assume that the acceleration due to gravity is approximately 32.2 ft/s².
Now, we can use the equation of motion for vertical motion:
h = h0 + v0*t - (1/2)*g*t²
Where:
h is the height of the rocket after time t
h0 is the initial height (ground level in this case, so h0 = 0)
v0 is the initial velocity (192 ft/s in this case)
g is the acceleration due to gravity (32.2 ft/s² in this case)
t is the time (10 s in this case)
Plugging in the given values into the equation, we get:
h = 0 + (192 ft/s)*(10 s) - (1/2)*(32.2 ft/s²)*(10 s)²
Simplifying this equation further:
h = 0 + 1920 ft - (1/2)*322 ft/s² * 100 s²
h = 1920 ft - 1610 ft
h = 310 ft
Therefore, the rocket is at a height of 310 feet after 10 seconds.