A stone is dropped from the upper observation deck of a tower, 200 m above the ground. (Assume g = 9.8 m/s^2.)
a)Find the distance (in meters) of the stone above ground level at time t.
b)With what velocity does it strike the ground? (Round your answer to one decimal place.)
c)If the stone is thrown downward with a speed of 2 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)
a = -g = -9.8
v = vi + a t = 0 -9.8 t
h = hi+vi t +.5 a t^2 = 200 +0- 4.9 t^2
there is an easy way to find v at ground using energy but this is calculus, not physics so:
0 = 200 - 4.9 t^2
t = 6.39 seconds to hit
v = 0 - 9.8 (.39 )
= -62.6 or 62.6 meters/second down
now if vi = -2
0 = 200 - 2t - 4.9 t^2
4.9 t^2 + 2 t - 200 = 0
solve with quadratic equation
a) Well, if you drop a stone from a height, it starts falling faster than an aspiring stand-up comedian's career. The equation to find the distance of the stone above the ground at time t is given by:
d = (1/2)gt^2 + h
Where d is the distance, g is acceleration due to gravity (approx. 9.8 m/s^2), t is time, and h is the initial height. Plugging in the values:
d = (1/2)(9.8)(t^2) + 200
b) Now let's talk velocity. The final velocity of the stone when it strikes the ground is like a bad punchline - no one likes it. But we can calculate it using the formula:
v = gt
Where v is the velocity and g is acceleration due to gravity (approx. 9.8 m/s^2). Plugging in the values:
v = (9.8)(t)
c) Finally, if you throw the stone downward with a speed of 2 m/s, it's like giving it an extra push, as if gravity wasn't doing enough already. We can find the time it takes to reach the ground using the equation:
0 = (1/2)gt^2 + vt + h
Where v is the initial velocity, g is acceleration due to gravity (approx. 9.8 m/s^2), t is time, and h is the initial height. Plugging in the values:
0 = (1/2)(9.8)(t^2) + (2)(t) + 200
You can solve these equations to find the answers, but remember, in the world of physics, timing is everything. Just like in comedy, timing is everything. Good luck!
a) The distance of the stone above the ground at time t can be calculated using the formula:
distance = initial height - (1/2) * g * t^2
Given:
Initial height (h) = 200 m
Acceleration due to gravity (g) = 9.8 m/s^2
Time (t) = ?
Substituting the values into the formula:
distance = 200 - (1/2) * 9.8 * t^2
distance = 200 - 4.9 * t^2
b) The velocity of the stone when it strikes the ground can be found using the formula:
velocity = initial velocity + g * t
Given:
Initial velocity (u) = 0 (since it is dropped)
Acceleration due to gravity (g) = 9.8 m/s^2
Time (t) = ?
Substituting the values into the formula:
velocity = 0 + 9.8 * t
velocity = 9.8 * t
c) To find the time it takes for the stone to reach the ground when thrown downward with a speed of 2 m/s, we can use the formula:
time = (final velocity - initial velocity) / acceleration
Given:
Initial velocity (u) = -2 m/s (downward)
Final velocity (v) = 0 (since it reaches the ground)
Acceleration due to gravity (g) = 9.8 m/s^2
Time (t) = ?
Substituting the values into the formula:
time = (0 - (-2)) / 9.8
time = 2 / 9.8
Let's calculate the values step by step.
a) The distance of the stone above ground level at time t:
distance = 200 - 4.9 * t^2
b) The velocity with which it strikes the ground:
velocity = 9.8 * t
c) The time it takes to reach the ground when thrown downward:
time = 2 / 9.8
To tackle these questions, we can use the kinematic equations of motion, specifically the equation for the position (displacement) and the equation for velocity.
a) The equation for position is given by:
s = s₀ + vt + (1/2)at²
In this case, the initial position (s₀) is 200 m (above ground level), the initial velocity (v₀) is 0 m/s (as the stone is dropped), the acceleration (a) is -9.8 m/s² (negative since it is due to gravity), and the time (t) is the variable we're solving for.
Rearranging the equation, we get:
s - s₀ = (1/2)at²
Substituting the known values:
0 - 200 = (1/2)(-9.8)t²
-200 = -4.9t²
Now solve for t by rearranging the equation:
t² = 200 / 4.9
t² ≈ 40.82
t ≈ √40.82
t ≈ 6.39 seconds
So, the stone will be above ground level at a distance of approximately 200 - (0.5)(9.8)(6.39²) = 10.06 meters after 6.39 seconds.
b) The equation for final velocity is given by:
v = v₀ + at
In this case, the initial velocity (v₀) is 0 m/s, the acceleration (a) is -9.8 m/s², and the time (t) is 6.39 seconds (from the previous part).
So, substituting the known values:
v = 0 + (-9.8)(6.39)
v ≈ -62.82 m/s (rounding to one decimal place)
Therefore, the stone will strike the ground with a velocity of approximately -62.8 m/s.
c) If the stone is thrown downward with an initial velocity (v₀) of 2 m/s, we can still use the equation for position to find the time it takes to reach the ground.
Using the same equation as before:
s - s₀ = (1/2)at²
Substituting the known values:
0 - 200 = (1/2)(-9.8)t²
-200 = -4.9t²
Rearranging the equation:
t² = 200 / 4.9
t² ≈ 40.82
t ≈ √40.82
t ≈ 6.39 seconds
Here, we can see that the time it takes for the stone to reach the ground is still approximately 6.39 seconds, regardless of the initial velocity.
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