Question
An object is dropped from the top of a cliff 610 meters high. Its height above the ground \( t \) seconds after it is dropped is \( 610-4.9 \mathrm{t}^{2} \). Determine its speed 10 seconds after it is dropped. The speed of the object 10 seconds after it is dropped is (Type an integer or a decimal. Do not round.)
Ask by Collins Tran. in the United States
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The speed of the object 10 seconds after it is dropped is 98 m/s.
Solution
1. The height of the object as a function of time is given by
\[
h(t) = 610 - 4.9t^2.
\]
2. The velocity is the derivative of the height with respect to time. Differentiating, we get
\[
v(t) = \frac{dh}{dt} = -9.8t.
\]
3. Evaluating the velocity at \( t = 10 \) seconds:
\[
v(10) = -9.8 \times 10 = -98 \text{ m/s}.
\]
4. The speed of the object is the magnitude of the velocity, so the speed is
\[
|v(10)| = 98 \text{ m/s}.
\]
The speed of the object 10 seconds after it is dropped is \( 98 \).
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Simplify this solution Mind Expander
To determine the speed of the object 10 seconds after it is dropped, we first find the object's position function, which is given by: \[ h(t) = 610 - 4.9t^2 \] To find the speed, we need to calculate the derivative of the height function with respect to time \( t \): \[ v(t) = \frac{dh}{dt} = -9.8t \] Now, substituting \( t = 10 \) seconds into the velocity function: \[ v(10) = -9.8(10) = -98 \, \text{m/s} \] The speed of the object 10 seconds after it is dropped is \( 98 \, \text{m/s} \) (the negative sign indicates the direction of motion is downward). So, the speed of the object 10 seconds after it is dropped is \( 98 \, \text{m/s} \).
