Question

You wad up a piece of paper and throw it into the wastebasket. How far will your piece of paper travel if you throw it with a velocity of \[4.3\text{ }m{{s}^{-1}}\]at an angle of \[65{}^\circ \]?

Answer 1

Hint: When a piece of paper is thrown upwards in the air, the paper is subjected to a projectile motion. In a projectile motion, there are two components of vectors. That is, the horizontal and vertical component. To solve the problems involving projectile motion, the kinematic equation is applied.

Complete answer:
Let V be the initial velocity, \[{{V}_{V}}\]be the initial vertical velocity, \[{{V}_{H}}\] be the initial horizontal velocity, A be the acceleration, t be the time taken, \[\theta \] be the angle made by the velocity vector with the horizontal and H be the height of projection.
As per the law of projectile motion and kinematics,
The height of projection is equal to the sum of the product of the initial velocity and the time when it is in the air, and 0.5 times the product of the acceleration constant and square of time.
It is expressed in the equation as,
\[H=Vt+0.5A{{t}^{2}}\]----(1)
Let this be equation (1)
When it is in the air, the velocity is the vertical component of velocity.
Therefore,
We use the vertical component of velocity in equation (1)
\[H={{V}_{V}}t+0.5A{{t}^{2}}\]---- (2)
Let this be equation (2)
In general,
The horizontal component of velocity vector is expressed by the equation
\[{{V}_{H}}=V\cos \theta \]--- (3)
Let this be equation (3)
Similarly,
The vertical component of velocity vector is expressed by the equation
\[{{V}_{V}}=V\sin \theta \]---- (4)
Let this be equation (4)
Substitute the values of V and θ in equation (4)
We get the vertical component of velocity vector as,
\[{{V}_{V}}=\text{ }4.3sin\left( 65 \right)\text{ }\]
After calculating, we get,
\[{{V}_{V}}=\text{ }3.9m{{s}^{-1}}\]--- (5)
Let this be equation (5)
When thrown upwards, acceleration acts towards the opposite direction.
Therefore, the value of acceleration when thrown upwards is expressed as,
\[A=-9.81m/{{s}^{2}}\]---(6)
Let this be equation (6)
Substitute the value of acceleration in equation (2)
We get,
\[H=3.9t-0.5\text{x}9.81{{t}^{2}}\]---(7)
Let this be equation (7)
In order to find t, we have to put H = 0
By putting H = 0 in equation (7),
We get,
\[H\text{ }=\text{ }0\]
\[3.9t-0.5\text{x}9.81{{t}^{2}}=0\]
 We get the values of t as,
\[t\text{ }=\text{ }0.4,\text{ }0\]
 \[t\text{ }=\text{ }0\] is impossible
But,
\[t\text{ }=\text{ }0.4s\]is possible.
Therefore, the answer is
\[t\text{ }=\text{ }0.4s\]

Note:
When an object moves from one point to another along a specified path, the particle is in projectile motion and the particle subjected to projectile motion is called a projectile. A projectile's motion is expressed in terms of its velocity, time and height. Examples of projectile motion includes a jet projecting in the air or a coin tossed in the air etc.