
An object is projected upwards with a velocity of \[100{\text{ }}m/s.\]It will strike the ground after (approximately)
A. \[10{\text{ }}sec\]
B. \[20{\text{ }}sec\]
C. \[15{\text{ }}sec\]
D. \[5{\text{ }}sec\]
Answer
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Hint:In order to calculate the time it takes for an object to fall in a given distance, we can use the formula \[t= \dfrac{v-u}{- g}\] where \[g\] is the standard gravity and \[t\] is time. In this problem, we have to calculate the time it takes for an object of mass \[100{\text{ }}kg\] to fall from a height of \[10{\text{ }}m.\]
Complete step by step solution:
In this experiment, we need to find the time when an object will strike the ground. To find the answer, we will do some simple math. When an object is projected upwards with a velocity of \[100{\text{ }}m/s,\] it takes only \[10{\text{ }}seconds\] for it to strike the ground.
We might want to consider how long it takes for gravity to bring the object back down. As gravity pulls an object down, we cannot predict when it will strike the ground again as it constantly moves around in a circle. This problem is related to the equation of motion. For our purposes, we will consider the object a cannonball shot vertically upwards from the ground.
After \[10{\text{ }}seconds\], it will have reached an altitude of \[1.73{\text{ }}meters\] above ground level and be on its way back down to earth. After \[20{\text{ }}seconds\], it will have reached an altitude of \[3.15{\text{ }}meters\] above ground level and be on its way back down to earth.
We need to figure out how long it takes to get to the ground so that we can figure out how long it takes to get there.
\[v{\text{ }} = {\text{ }}u{\text{ }} - {\text{ }}gt\;,\] Implies:
\[t= \dfrac{v-u}{- g} \\
\Rightarrow t = \dfrac{0 - 100}{- 10t} \\
\therefore t = {\text{ }}10{\text{ }}seconds{\text{ }}. \\ \]
As a result, the time it takes to reach the earth is about is: \[10{\text{ }}s{\text{ }} + {\text{ }}10s{\text{ }} = {\text{ }}20s{\text{ }}seconds.\]
Hence, option B is the correct answer.
Note: The velocity calculates the time it takes for the object to strike the ground. The velocity calculates the time it takes for an object with a given velocity to reach a target point.
Complete step by step solution:
In this experiment, we need to find the time when an object will strike the ground. To find the answer, we will do some simple math. When an object is projected upwards with a velocity of \[100{\text{ }}m/s,\] it takes only \[10{\text{ }}seconds\] for it to strike the ground.
We might want to consider how long it takes for gravity to bring the object back down. As gravity pulls an object down, we cannot predict when it will strike the ground again as it constantly moves around in a circle. This problem is related to the equation of motion. For our purposes, we will consider the object a cannonball shot vertically upwards from the ground.
After \[10{\text{ }}seconds\], it will have reached an altitude of \[1.73{\text{ }}meters\] above ground level and be on its way back down to earth. After \[20{\text{ }}seconds\], it will have reached an altitude of \[3.15{\text{ }}meters\] above ground level and be on its way back down to earth.
We need to figure out how long it takes to get to the ground so that we can figure out how long it takes to get there.
\[v{\text{ }} = {\text{ }}u{\text{ }} - {\text{ }}gt\;,\] Implies:
\[t= \dfrac{v-u}{- g} \\
\Rightarrow t = \dfrac{0 - 100}{- 10t} \\
\therefore t = {\text{ }}10{\text{ }}seconds{\text{ }}. \\ \]
As a result, the time it takes to reach the earth is about is: \[10{\text{ }}s{\text{ }} + {\text{ }}10s{\text{ }} = {\text{ }}20s{\text{ }}seconds.\]
Hence, option B is the correct answer.
Note: The velocity calculates the time it takes for the object to strike the ground. The velocity calculates the time it takes for an object with a given velocity to reach a target point.
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