Questions: When an object is dropped on a certain earth-like planet, the distance it falls in t seconds, assuming that air resistance is negligible, is given by s(t) = 19t^2 where s(t) is in feet. Suppose that a medic's reflex hammer is dropped from a hovering helicopter. Find (a) how far the hammer falls in 4 sec, (b) how fast the hammer is traveling 4 sec after being dropped, and (c) the hammer's acceleration after it has been falling for 4 sec. (a) The hammer falls 304 feet in 4 seconds. (Simplify your answer.) (b) The hammer is traveling [] ft/sec 4 seconds after being dropped. (Simplify your answer.)

When an object is dropped on a certain earth-like planet, the distance it falls in t seconds, assuming that air resistance is negligible, is given by s(t) = 19t^2 where s(t) is in feet. Suppose that a medic's reflex hammer is dropped from a hovering helicopter. Find (a) how far the hammer falls in 4 sec, (b) how fast the hammer is traveling 4 sec after being dropped, and (c) the hammer's acceleration after it has been falling for 4 sec. (a) The hammer falls 304 feet in 4 seconds. (Simplify your answer.) (b) The hammer is traveling [] ft/sec 4 seconds after being dropped. (Simplify your answer.)
Transcript text: When an object is dropped on a certain earth-like planet, the distance it falls in t seconds, assuming that air resistance is negligible, is given by $s(t) = 19t^2$ where s(t) is in feet. Suppose that a medic's reflex hammer is dropped from a hovering helicopter. Find (a) how far the hammer falls in 4 sec, (b) how fast the hammer is traveling 4 sec after being dropped, and (c) the hammer's acceleration after it has been falling for 4 sec. (a) The hammer falls 304 feet in 4 seconds. (Simplify your answer.) (b) The hammer is traveling [] ft/sec 4 seconds after being dropped. (Simplify your answer.)
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Solution

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Solution Steps

Step 1: Calculate the Distance Fallen

To find the distance an object has fallen after 4 seconds, we use the equation $s(t) = k t^2$. Substituting the given values, $s(4) = 19 \times 4^2 = 304$ units.

Step 2: Calculate the Velocity

The velocity of the object at any time $t$ can be found by differentiating the distance equation with respect to time. This gives $v(t) = 2kt = 2 \times 19 \times 4 = 152$ units per second.

Step 3: Calculate the Acceleration

The acceleration of the object is constant and can be found by differentiating the velocity equation with respect to time or directly from the knowledge that $a = 2k$. Thus, the acceleration due to gravity on this planet is $a = 2 \times 19 = 38$ units per square second.

Final Answer:

The distance fallen after 4 seconds is 304 units, the velocity at 4 seconds is 152 units per second, and the constant acceleration due to gravity is 38 units per square second.

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