Question

If a body is projected with a velocity of 9.8 $m{s^{ - 1}}$ making an angle of ${45^o}$with the horizontal, then the range of the projectile is (take $g{\text{ }} = {\text{ }}9.8{\text{ }}m{s^{ - 2}}$)
(A) 39.2 m
(B) 9.8 m
(C) 4.9 m
(D) 19.6 m

Answer 1

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:

{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