Anobject is dropped from 320m high. Find the time of motion and velocity when it hits the ground. Assume gravity is 9.8 meters per second squires.
the height is h(t) = 320 - 4.9t^2, right?
So find t when h=0
and of course, v(t) = -9.8t
To find the time of motion and velocity when an object hits the ground after being dropped from a certain height, we can use the equations of motion. In this case, we can use the equation for vertical motion, which is given by:
s = ut + (1/2)at^2
Where:
- s is the displacement (distance) traveled by the object
- u is the initial velocity (which is zero as the object is dropped)
- t is the time taken
- a is the acceleration (which is equal to gravity, -9.8 m/s^2)
We need to find the time of motion (t) when the object hits the ground. The displacement (s) in this case is equal to 320 m, and the initial velocity (u) is zero.
Plugging these values into the equation, we get:
320 = 0*t + (1/2)*(-9.8)*t^2
Simplifying the equation, we have:
320 = -4.9*t^2
Rearranging the equation, we get:
t^2 = -320 / -4.9
t^2 = 65.3
Taking the square root of both sides, we find:
t = √(65.3)
t ≈ 8.08 seconds (rounded to two decimal places)
So, the time of motion when the object hits the ground is approximately 8.08 seconds.
To find the velocity when it hits the ground, we can use the equation for velocity in vertical motion, given by:
v = u + at
Since the initial velocity (u) is zero, the equation simplifies to:
v = at
Substituting the values, we have:
v = -9.8 * 8.08
v ≈ -79.4 m/s (rounded to one decimal place)
Therefore, when the object hits the ground, its velocity would be approximately -79.4 m/s. The negative sign indicates that the velocity is directed downward.