a driver runs off the end of a 6.50 m tall diving board with an initial horizontal velocity of 7.05 m/s.
How long is the diver in the air?
How far away from the cliff does the diver land?
What is the final vertical velocity just before the diver lands?
flight time (t) ... 6.50 = 1/2 g t^2
horizontal distance ... 7.05 * t
vert vel at impact ... g * t ... downward
To find the time the diver is in the air, we can use the vertical motion equation:
h = ut + (1/2)gt^2,
where:
- h is the initial height of the diving board (6.50 m),
- u is the initial vertical velocity (0 m/s as the diver jumps horizontally),
- g is the acceleration due to gravity (-9.8 m/s^2),
- t is the time the diver is in the air (what we're trying to find).
Plugging in the values into the equation, we get:
6.50 = 0 + (1/2)(-9.8)t^2.
Simplifying this equation, we get:
6.50 = -4.9t^2.
To solve for t, we rearrange the equation as:
t^2 = -6.50 / -4.9,
t^2 = 1.3265.
Taking the square root of both sides to solve for t, we get:
t = √1.3265.
Calculating this, we find that time t ≈ 1.152 seconds.
Now let's find how far away from the cliff the diver lands. Since the initial horizontal velocity of the diver is 7.05 m/s, we can use the formula:
d = ut,
where:
- d is the distance traveled (what we're trying to find),
- u is the initial horizontal velocity (7.05 m/s),
- t is the time (1.152 seconds).
Plugging in the values:
d = (7.05)(1.152),
Calculating this, we find that the diver lands approximately 8.11 meters away from the cliff.
To find the final vertical velocity just before the diver lands, we can use the equation:
v = u + gt,
where:
- v is the final vertical velocity (what we're trying to find),
- u is the initial vertical velocity (0 m/s as the diver jumps horizontally),
- g is the acceleration due to gravity (-9.8 m/s^2),
- t is the time (1.152 seconds).
Plugging in the values:
v = 0 + (-9.8)(1.152),
Calculating this, we find that the final vertical velocity just before the diver lands is approximately -11.26 m/s (assuming downward as negative).