A projectile is fired straight up from a cliff which is 200 feet above ground with an initial velocity of 220 ft/sec. Assume that the only force affecting the projectile during flight is from gravity which produces a downward acceleration 32 ft/sec^2. Find an equation from the projectile's height above the ground as a function of time t ( if t = 0 when the projectile is fired).
LOL - look above where you called it a Physics question.
The position function would be
s(t)=-32t^2+220t+200
Enjoy! Sorry i replied 4 years late because this the only time i saw a website like this.
It -16t^2 not -32t^2 sorry for the wrong info i was so sleepy that time. :) good luck
To find the equation for the projectile's height above the ground as a function of time, we need to consider the motion equations for the vertical direction.
The general equation for motion in the vertical direction (assuming constant acceleration) is given by:
h(t) = h_0 + v_0 * t + (1/2) * a * t^2
where:
h(t) is the height above the ground at time t,
h_0 is the initial height (200 ft),
v_0 is the initial velocity (220 ft/sec),
a is the acceleration (-32 ft/sec^2),
and t is the time elapsed since the projectile was fired.
Given that the initial height (h_0) is 200 ft, the initial velocity (v_0) is 220 ft/sec, and the downward acceleration (a) is -32 ft/sec^2, we can substitute these values into the equation:
h(t) = 200 + 220 * t + (1/2) * (-32) * t^2
Simplifying the equation further, we get:
h(t) = 200 + 220t - 16t^2
So, the equation for the projectile's height above the ground as a function of time (t) is h(t) = 200 + 220t - 16t^2.