Respuesta :
d = 950 m - 4.9t^2 m
The distance an object moves under constant acceleration is
d = 0.5at^2
where
d = distance
a = acceleration
t = time.
Since we're falling and since we're starting at 950 m above ground, the formula becomes:
d = 950 m - 0.5at^2
Substituting known values, and simplifying gives us
d = 950 m - 0.5*9.8 m/s^2 * t^2
d = 950 m - 4.9 m/s^2 * t^2
Since time is in seconds, we can cancel out the seconds in the units, getting
d = 950 m - 4.9t^2 m
The distance of the stone above the ground at time [tex]t[/tex] will be given by [tex]\boxed{d=\left( {950 - 4.9{t^2}}\right)\,{\text{m}}}[/tex]
Further Explanation:
Given:
The stone is dropped from a height of [tex]950\,{\text{m}}[/tex].
The acceleration due to gravity is [tex]9.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex].
Concept:
Since the stone is dropped, the initial velocity of the stone will be zero.
The distance covered by the stone as it falls under gravity for time [tex]t[/tex] is give by the second equation of motion. The equation of motion is written as:
[tex]s= ut+\dfrac{1}{2}g{t^2}[/tex]
Here, [tex]s[/tex] is the distance covered by the stone in time [tex]t[/tex], [tex]u[/tex] is the initial velocity of the stone and [tex]g[/tex] is the acceleration due to gravity on the surface of Earth.
Substitute the value of initial velocity and the acceleration due to gravity in equation.
[tex]\begin{aligned}s &= \left( {0 \times t} \right) + \left( {\frac{1}{2} \times 9.8 \times {t^2}} \right) \\&=4.9{t^2}\\\end{aligned}[/tex]
This is the distance covered by the stone as it falls. The distance of the stone from the ground is given by:
[tex]d = D - s[/tex]
Here, [tex]D[/tex] is the total height from which the stone falls and [tex]d[/tex] is the height of the stone from the ground.
Substitute the value of [tex]D[/tex] and [tex]s[/tex] in above expression.
[tex]\boxed{d=950-4.9{t^2}}[/tex]
Thus, the distance of the stone above the ground at time [tex]t[/tex] will be given by
[tex]\boxed{d = \left( {950 - 4.9{t^2}} \right)\,{\text{m}}}[/tex]
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Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords: Stone, dropped, under observation deck, 950m above the ground, acceleration due to gravity, distance of the stone, at time t, initial velocity.
