A coconut is hanging on a tree at a height of 15 m from the ground. A body launches a projectile vertically upwards with a velocity of 20 m/s. After what time will the projectile pass by the coconut. Explain the two answers you get in this problem.
Answer
614.4k+ views
Hint: The above problem can be resolved using the kinematic equations of motion. The Kinematic equations of motions relate the speed and distance undertaken for the given interval of time.
Complete step by step answer:
Given:
The height of tree is, \[s = 15\;{\rm{m}}\]
The initial speed is, \[u = 20\;{\rm{m/s}}\].
Applying the second kinematic equation of motion as
\[s = ut + \dfrac{1}{2}g{t^2}\]
Here, t is the time and g is the gravitational acceleration and its value is \[9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\].
Solving by substituting the values as,
\[\begin{array}{l}
s = ut + \dfrac{1}{2}g{t^2}\\
15\;{\rm{m}} = 20\;{\rm{m/s}} \times t + \dfrac{1}{2} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \times {t^2}\\
3 = 4t - {t^2}
\end{array}\]
Solving the quadratic equation as,
\[t = 1,3\]
Therefore, at time t=1 second the projectile passes the coconut towards upwards direction and at t=3 seconds it passes the coconut in downward direction.
Note:
To resolve the given problem, the proper analysis of kinematic equations of motion is required to be done. The necessary points like initial speed, the height and time are to be pointed out from the given condition, to frame out the suitable kinematic equation of motion. Moreover, the conclusion is to be on point and carry some significance with respect to the given problem.
Complete step by step answer:
Given:
The height of tree is, \[s = 15\;{\rm{m}}\]
The initial speed is, \[u = 20\;{\rm{m/s}}\].
Applying the second kinematic equation of motion as
\[s = ut + \dfrac{1}{2}g{t^2}\]
Here, t is the time and g is the gravitational acceleration and its value is \[9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\].
Solving by substituting the values as,
\[\begin{array}{l}
s = ut + \dfrac{1}{2}g{t^2}\\
15\;{\rm{m}} = 20\;{\rm{m/s}} \times t + \dfrac{1}{2} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \times {t^2}\\
3 = 4t - {t^2}
\end{array}\]
Solving the quadratic equation as,
\[t = 1,3\]
Therefore, at time t=1 second the projectile passes the coconut towards upwards direction and at t=3 seconds it passes the coconut in downward direction.
Note:
To resolve the given problem, the proper analysis of kinematic equations of motion is required to be done. The necessary points like initial speed, the height and time are to be pointed out from the given condition, to frame out the suitable kinematic equation of motion. Moreover, the conclusion is to be on point and carry some significance with respect to the given problem.
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