Sounds like a math textbook question. Physics books use metric units :)
for constant acceleration:
x = Xi + Vi t + (1/2) a t^2
so here
h = 200 + 220 t - 16 t^2
for constant acceleration:
x = Xi + Vi t + (1/2) a t^2
so here
h = 200 + 220 t - 16 t^2
h = h0 + v0 * t - (1/2) * g * t^2
where:
h is the final height above the ground,
h0 is the initial height above the ground,
v0 is the initial velocity,
g is the acceleration due to gravity, and
t is the time elapsed.
Given:
h0 = 200 feet (initial height),
v0 = 220 ft/sec (initial velocity),
g = 32 ft/sec^2 (acceleration due to gravity).
Substituting these values into the equation, we get:
h = 200 + 220 * t - (1/2) * 32 * t^2
Simplifying further:
h = 200 + 220t - 16t^2
Therefore, the equation for the projectile's height above the ground as a function of time t is h = 200 + 220t - 16t^2.
h(t) = h0 + v0t + (1/2)gt^2
Where:
- h(t) is the height of the projectile above the ground at time t
- h0 is the initial height of the projectile
- v0 is the initial velocity of the projectile
- g is the acceleration due to gravity
- t is the time elapsed since the projectile was fired
Given that the cliff is 200 feet above the ground, the initial height (h0) is 200 feet, the initial velocity (v0) is 220 ft/sec, and the acceleration due to gravity (g) is -32 ft/sec^2 (negative because it's acting in the opposite direction of the positive axis), we can substitute these values into the equation:
h(t) = 200 + 220t - (1/2)32t^2
Simplifying, we get the final equation:
h(t) = 200 + 220t - 16t^2