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Question: A boy of height 1.5 m, making move on a skateboard due east with velocity \(4m{s^{ - 1}}\), throws a...

A boy of height 1.5 m, making move on a skateboard due east with velocity 4ms−14m{s^{ - 1}}, throws a coin vertically up with a velocity of 3ms−13m{s^{ - 1}} relative to himself. Find the total displacement of the coin relative to ground till it comes to the hand of the boy. What is the maximum height attained by the coin w.r.t to ground?

Explanation

Solution

Displacement is defined as the shortest distance between the initial and final point of the motion. The acceleration acting on the body which is thrown vertically up is acceleration due to gravity and the direction of the acceleration is in downwards.

Formula used:
The formula of the second relation of Newton’s law of motion is given by,
⇒s⃗=u⃗t+12a⃗t2\Rightarrow \vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}
Where the displacement vector is s⃗\vec s, the initial velocity vector is u⃗\vec u, the acceleration vector is a⃗\vec a and the time taken is t.
The formula of the third relation of Newton’s law of motion is given by,
⇒v⃗2−u⃗2=2⋅a⃗⋅s⃗\Rightarrow {\vec v^2} - {\vec u^2} = 2 \cdot \vec a \cdot \vec s
Where the initial velocity vector is u⃗\vec u, the final velocity vector is v⃗\vec v, the acceleration vector is a⃗\vec a and the displacement vector is s⃗\vec s.

Complete step by step solution:
It is given in the problem that a boy 1.5 m having height making move on skateboard due east with velocity 4ms−14m{s^{ - 1}} and he throws a coin vertically up with a velocity of 3ms−13m{s^{ - 1}} relative to himself, we need to tell the total displacement and the maximum height that the coin attains with respect to the coin.
The velocity of the coin is equal to,
⇒v⃗coin=v⃗coin,boy+v⃗boy,ground\Rightarrow {\vec v_{coin}} = {\vec v_{coin,boy}} + {\vec v_{boy,ground}}
⇒v⃗coin=4i^+3j^\Rightarrow {\vec v_{coin}} = 4\hat i + 3\hat j
The acceleration on the coin is equal toa⃗=−10k^ ms−2\vec a = - 10\hat k{\text{ }}m{s^{ - 2}}.
The displacement of the coin in the vertical direction is zero as the coin returns to its initial position.
The formula of the second relation of Newton’s law of motion is given by,
⇒s⃗=u⃗t+12a⃗t2\Rightarrow \vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}
Where the displacement vector is s⃗\vec s, the initial velocity vector is u⃗\vec u, the acceleration vector is a⃗\vec a and the time taken is t.
For motion in vertical direction.
The displacement is zero the initial velocity is 3ms−13m{s^{ - 1}} the acceleration on the coin is g=−10ms2g = - 10\dfrac{m}{{{s^2}}} and let the time taken for the journey is t.
⇒s=ut+12at2\Rightarrow s = ut + \dfrac{1}{2}a{t^2}
⇒0=3t−12(10)t2\Rightarrow 0 = 3t - \dfrac{1}{2}\left( {10} \right){t^2}
⇒0=3t−5t2\Rightarrow 0 = 3t - 5{t^2}
⇒t(3−5t)=0\Rightarrow t\left( {3 - 5t} \right) = 0
So the time taken is equal to t=35sec⁡.t = \dfrac{3}{5}\sec ..
The displacement of the coin in the x-direction is equal to,
⇒d=v×t\Rightarrow d = v \times t
⇒d=4×35\Rightarrow d = 4 \times \dfrac{3}{5}
⇒d=125m\Rightarrow d = \dfrac{{12}}{5}m.
The horizontal displacement is equal to d=125md = \dfrac{{12}}{5}m.
Let us calculate the maximum height that the coin reaches in the vertical direction.
The formula of the third relation of Newton’s law of motion is given by,
⇒v⃗2−u⃗2=2⋅a⃗⋅s⃗\Rightarrow {\vec v^2} - {\vec u^2} = 2 \cdot \vec a \cdot \vec s
Where the initial velocity vector is u⃗\vec u, the final velocity vector isv⃗\vec v, the acceleration vector is a⃗\vec a and the displacement vector iss⃗\vec s.
⇒v2−u2=2as\Rightarrow {v^2} - {u^2} = 2as
The final velocity is zero at the highest point the initial velocity is 3ms−13m{s^{ - 1}} the acceleration is g=−10ms2g = - 10\dfrac{m}{{{s^2}}} and let the height is equal to h.
⇒v2−u2=2as\Rightarrow {v^2} - {u^2} = 2as
⇒0−32=2×(−10)×(h)\Rightarrow 0 - {3^2} = 2 \times \left( { - 10} \right) \times \left( h \right)
⇒32=2×(10)×(h)\Rightarrow {3^2} = 2 \times \left( {10} \right) \times \left( h \right)
⇒9=20×(h)\Rightarrow 9 = 20 \times \left( h \right)
⇒h=920m\Rightarrow h = \dfrac{9}{{20}}m
The maximum height that the coin will reach is equal to,
⇒H=(h+1⋅5)m\Rightarrow H = \left( {h + 1 \cdot 5} \right)m
⇒H=(920+1⋅5)m\Rightarrow H = \left( {\dfrac{9}{{20}} + 1 \cdot 5} \right)m
⇒H=1⋅95m\Rightarrow H = 1 \cdot 95m

The maximum height reached by the coin is equal to H=1⋅95mH = 1 \cdot 95m.

Note: The final velocity of the body when a body is thrown upwards becomes zero at the highest point of the motion and then the body starts coming downwards and at the highest point of the motion the velocity of the body is zero but acceleration on the body is equal to acceleration due to gravity.