Answer
Verified
89.4k+ views
Hint: The range of the projectile is the max horizontal distance covered by the projectile. This is calculated using the expression relating the projectile velocity, acceleration due to gravity and angle of projection. Substitute the values in S.I units to find the range.
Complete step-by-step answer
For a projectile motion the horizontal distance traveled by a body during the time of flight is range. Assuming that the staring and the end point are at equal height, the range of the projectile (R) is given by
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Where,
u is the velocity of the object
θ is the angle of projection
g is the acceleration due to gravity
The data given in the problem:
\[\begin{array}{*{20}{l}}
{u{\text{ }} = {\text{ }}9.8{\text{ }}m{s^{ - 1}}} \\
{g{\text{ }} = {\text{ }}9.8{\text{ }}m{s^{ - 2}}} \\
{\theta = {\text{ }}{{45}^0}}
\end{array}\]
Now substitute the known data in the range formula
$
R = \dfrac{{{{9.8}^2} \times \sin 90}}{{9.8}} \\
R = 9.8m \\
$
Hence, the range of the projectile is 9.8 m and the correct option is B
Note
1. Range of the projectile is max when it is projected with an angle of 45o.
Range of the projectile can also be expressed as
$R = \dfrac{{2{u_x}{u_y}}}{g}$
Where, $u_x$ and $u_y$ are the components of initial velocity.
2. In a general case: the starting and the end point of the projectile need not be at the same height.
Then this formula can be used
$R = \dfrac{{{v^2}}}{{2g}}\left( {1 + \sqrt {1 + \dfrac{{2gy}}{{2{{\sin }^2}\theta }}} } \right)$
Here R is the range, v is the launching velocity, g is the acceleration due to gravity, y is the difference in height between the starting and ending point from the ground, $\theta $ is the initial angle of launch
Complete step-by-step answer
For a projectile motion the horizontal distance traveled by a body during the time of flight is range. Assuming that the staring and the end point are at equal height, the range of the projectile (R) is given by
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Where,
u is the velocity of the object
θ is the angle of projection
g is the acceleration due to gravity
The data given in the problem:
\[\begin{array}{*{20}{l}}
{u{\text{ }} = {\text{ }}9.8{\text{ }}m{s^{ - 1}}} \\
{g{\text{ }} = {\text{ }}9.8{\text{ }}m{s^{ - 2}}} \\
{\theta = {\text{ }}{{45}^0}}
\end{array}\]
Now substitute the known data in the range formula
$
R = \dfrac{{{{9.8}^2} \times \sin 90}}{{9.8}} \\
R = 9.8m \\
$
Hence, the range of the projectile is 9.8 m and the correct option is B
Note
1. Range of the projectile is max when it is projected with an angle of 45o.
Range of the projectile can also be expressed as
$R = \dfrac{{2{u_x}{u_y}}}{g}$
Where, $u_x$ and $u_y$ are the components of initial velocity.
2. In a general case: the starting and the end point of the projectile need not be at the same height.
Then this formula can be used
$R = \dfrac{{{v^2}}}{{2g}}\left( {1 + \sqrt {1 + \dfrac{{2gy}}{{2{{\sin }^2}\theta }}} } \right)$
Here R is the range, v is the launching velocity, g is the acceleration due to gravity, y is the difference in height between the starting and ending point from the ground, $\theta $ is the initial angle of launch
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The diagram given shows how the net interaction force class 11 physics JEE_Main
Ethylidene dichloride is obtained by the reaction of class 12 chemistry JEE_Main
Assertion The trajectory of a projectile is quadratic class 11 physics JEE_Main
Which of the following meet the requirements of Huckel class 11 chemistry JEE_Main
A block of mass 5 kg is on a rough horizontal surface class 11 physics JEE_Main
Choose the incorrect statement from the following A class 12 physics JEE_Main