
The horizontal and vertical components of the initial velocity of a projectile are \[{\text{10}}\,{\text{m/s}}\]and ${\text{20}}\,{\text{m/s}}$respectively. The horizontal range of the projectile will be $\left[ {g\, = \,10\,m/{s^2}} \right]$:-
A. ${\text{5m}}$
B. ${\text{10m}}$
C. ${\text{20m}}$
D. ${\text{40m}}$
Answer
522.3k+ views
Hint:Any particle thrown by the force applied is referred to as a projectile. It can also be described as an object that is launched into space and permitted to move freely due to gravity and air resistance. While projectiles can be any object in motion through space (such as a thrown baseball, kicked football), they are most commonly used in warfare and sports.
Complete step by step answer:
Owing to gravity's impact, projectiles follow a parabolic trajectory. Gravity causes a vertical acceleration of ${\text{9}}{\text{.8}}\,{\text{m/s}}$, which is downward. Every second, a projectile's vertical velocity varies by $9.8$ metres per second. A projectile's horizontal motion is independent of its vertical motion. We are given that,
The horizontal component of the initial velocity of a projectile,
$\left( {{{vcos\theta }}} \right)\,\, = \,\,{\text{10}}\,{\text{m/s}}$
The vertical component of the initial velocity of a projectile,
$\left( {{{vsin\theta }}} \right)\,\,{\text{ = }}\,\,{\text{20}}\,{\text{m/s}}$
The horizontal range of the projectile is given by
${\text{R}}\,\,{\text{ = }}\,\,\dfrac{{{\text{2}}\,{{ \times vsin\theta \times cos\theta }}}}{{{\text{10}}}}$
Substituting the values in the above equation, we get
$R\,\, = \,\,\dfrac{{2 \times 20 \times 10}}{{10}}$
On solving the above equations, we get
$\therefore {\text{R}}\,\,{\text{ = }}\,\,{\text{40m}}$
Hence, the horizontal range of the projectile will be ${\text{40}}\,{\text{m}}$.
So, the correct option is D.
Note:Projectile motion has a wide range of uses in physics and engineering. Meteors entering the Earth's atmosphere, fireworks, and the motion of any ball in sports are all examples. Projectiles are such objects, and their direction is referred to as a trajectory. The motion of falling objects is a straightforward one-dimensional projectile motion with no horizontal movement.
Complete step by step answer:
Owing to gravity's impact, projectiles follow a parabolic trajectory. Gravity causes a vertical acceleration of ${\text{9}}{\text{.8}}\,{\text{m/s}}$, which is downward. Every second, a projectile's vertical velocity varies by $9.8$ metres per second. A projectile's horizontal motion is independent of its vertical motion. We are given that,
The horizontal component of the initial velocity of a projectile,
$\left( {{{vcos\theta }}} \right)\,\, = \,\,{\text{10}}\,{\text{m/s}}$
The vertical component of the initial velocity of a projectile,
$\left( {{{vsin\theta }}} \right)\,\,{\text{ = }}\,\,{\text{20}}\,{\text{m/s}}$
The horizontal range of the projectile is given by
${\text{R}}\,\,{\text{ = }}\,\,\dfrac{{{\text{2}}\,{{ \times vsin\theta \times cos\theta }}}}{{{\text{10}}}}$
Substituting the values in the above equation, we get
$R\,\, = \,\,\dfrac{{2 \times 20 \times 10}}{{10}}$
On solving the above equations, we get
$\therefore {\text{R}}\,\,{\text{ = }}\,\,{\text{40m}}$
Hence, the horizontal range of the projectile will be ${\text{40}}\,{\text{m}}$.
So, the correct option is D.
Note:Projectile motion has a wide range of uses in physics and engineering. Meteors entering the Earth's atmosphere, fireworks, and the motion of any ball in sports are all examples. Projectiles are such objects, and their direction is referred to as a trajectory. The motion of falling objects is a straightforward one-dimensional projectile motion with no horizontal movement.
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