Question

A stone is dropped from the upper observation deck of a tower, 950 m above the ground. (assume g = 9.8 m/s2.) (a) find the distance (in meters) of the stone above ground level at time t.

Answer 1

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

Answer 2

The distance of the stone above the ground at time $t$ will be given by  $\boxed{d=\left( {950 - 4.9{t^2}}\right)\,{\text{m}}}$

Further Explanation:

Given:

The stone is dropped from a height of  $950\,{\text{m}}$.

The acceleration due to gravity is  $9.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}$.

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 $t$ is give by the second equation of motion. The equation of motion is written as:

$s= ut+\dfrac{1}{2}g{t^2}$

Here, $s$ is the distance covered by the stone in time $t$, $u$ is the initial velocity of the stone and $g$ 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.

$\begin{aligned}s &= \left( {0 \times t} \right) + \left( {\frac{1}{2} \times 9.8 \times {t^2}} \right) \\&=4.9{t^2}\\\end{aligned}$

This is the distance covered by the stone as it falls. The distance of the stone from the ground is given by:

$d = D - s$

Here, $D$ is the total height from which the stone falls and $d$ is the height of the stone from the ground.

Substitute the value of $D$ and $s$ in above expression.

$\boxed{d=950-4.9{t^2}}$

Thus, the distance of the stone above the ground at time $t$ will be given by

$\boxed{d = \left( {950 - 4.9{t^2}} \right)\,{\text{m}}}$

Learn More:

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3. A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high brainly.com/question/161035

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.