An object is dropped from 320m high. Find the time of motion and velocity when it hits the ground.
Ignoring air friction?
h=vo*t+1/2 g t^2
320=0-4.8 t^2 solve for time of flight t.
velocity= vo+gt but g=-9.8 m/s^2, you have t, solve for velocity. Vo is again zero.
To solve this problem, we can use the equations of motion for objects in free fall. We'll use the following equations:
1. h = ut + (1/2)gt^2
2. v = u + gt
Where:
- h is the height of the object
- u is the initial velocity (which in this case is 0 since the object is dropped)
- g is the acceleration due to gravity, which is approximately 9.8 m/s²
- t is the time taken
- v is the final velocity (the velocity when it hits the ground)
First, we'll use the equation for height (h) to find the time of motion (t). Rearranging equation 1, we have:
h = (1/2)gt^2
Plugging in the given values, we have:
320 = (1/2)(9.8)t^2
Now we can solve for t. Rearrange the equation and divide both sides by (1/2)(9.8):
320 / [(1/2)(9.8)] = t^2
Simplifying:
t^2 = 320 / [(1/2)(9.8)]
t^2 = 320 / 4.9
t^2 = 65.31
Taking the square root of both sides:
t ≈ 8.09 seconds
So, the time it takes for the object to hit the ground is approximately 8.09 seconds.
Now, let's find the final velocity (v) when the object hits the ground. We'll use equation 2:
v = u + gt
Since the object is dropped, the initial velocity (u) is 0, so we have:
v = 0 + (9.8)(8.09)
v ≈ 79.32 m/s
Therefore, the velocity when the object hits the ground is approximately 79.32 m/s.