
A bullet is fired with a speed of $1500m/s$ in order to hit a target \[100m\] away. If $g=10m/s^{2}$ the gun should be aimed:
\[\begin{align}
& \text{A}\text{.5cm }\!\!~\!\!\text{ above the target} \\
& \text{B}\text{. }\!\!~\!\!\text{ 10cm above the target} \\
& \text{C}\text{. }\!\!~\!\!\text{ 2}\text{.2cm above the target} \\
& \text{D}\text{. directly towards the target}\text{.} \\
\end{align}\]
Answer
513k+ views
Hint: This sum is an example of projectile motion, where the range$R=100m$, to calculate the height of the motion, or the distance along the $S_{y}$, we can use the following $S_{y}=\dfrac{1}{2}gt^{2}$. To find the time $t$taken to hit the target, we can use$v=\dfrac{R}{t}$.
Formulas used:
$v=\dfrac{R}{t}$and $S_{y}=\dfrac{1}{2}gt^{2}$
Complete step by step answer:
Here, the bullet follows projectile motion, to reach the target, which is at a distance $R=100m$. Then we know that the speed of the bullet $v=\dfrac{R}{t}$, where $t$ is the time taken.
Then, the time taken to hit the target is given as $t=\dfrac{100}{1500}=\dfrac{1}{15}s$
Since, the bullet travels along the x-axis and y-axis during this time $t$. The maximum height to which the bullet travels is given by $S_{y}=\dfrac{1}{2}gt^{2}$.So we aim the gun along the y-axis, and then the distance along the y-axis is given by $S_{y}=\dfrac{1}{2}gt^{2}$.
We just found that the\[\], then $S_{y}=\dfrac{1}{2}\times 10\times (\dfrac{1}{15})^{2}=0.022m=2.2cm$
Hence, to attack the target, the \[2.2cm\] above the target.
Hence the answer is option. \[2.2cm\] above the target.
Additional Information:
A body experiences projectile motion, when it is projected at an angle above the earth’s surface. This body retraces a curved path called the ballistic trajectory under gravity, before touching the ground. The maximum height the body moves is given by $S_{y}=\dfrac{1}{2}gt^{2}$ during the time taken $t$ .
Note:
Here, the gun is aimed at a certain angle above the ground. Without the details of the angle we can still find the height the bullet reaches, before hitting the target. Since the bullet starts from the gun at a certain time, it travels both along the x-axis and the y-axis. We use this logic to solve the sum.
Formulas used:
$v=\dfrac{R}{t}$and $S_{y}=\dfrac{1}{2}gt^{2}$
Complete step by step answer:
Here, the bullet follows projectile motion, to reach the target, which is at a distance $R=100m$. Then we know that the speed of the bullet $v=\dfrac{R}{t}$, where $t$ is the time taken.
Then, the time taken to hit the target is given as $t=\dfrac{100}{1500}=\dfrac{1}{15}s$
Since, the bullet travels along the x-axis and y-axis during this time $t$. The maximum height to which the bullet travels is given by $S_{y}=\dfrac{1}{2}gt^{2}$.So we aim the gun along the y-axis, and then the distance along the y-axis is given by $S_{y}=\dfrac{1}{2}gt^{2}$.
We just found that the\[\], then $S_{y}=\dfrac{1}{2}\times 10\times (\dfrac{1}{15})^{2}=0.022m=2.2cm$
Hence, to attack the target, the \[2.2cm\] above the target.
Hence the answer is option. \[2.2cm\] above the target.
Additional Information:
A body experiences projectile motion, when it is projected at an angle above the earth’s surface. This body retraces a curved path called the ballistic trajectory under gravity, before touching the ground. The maximum height the body moves is given by $S_{y}=\dfrac{1}{2}gt^{2}$ during the time taken $t$ .
Note:
Here, the gun is aimed at a certain angle above the ground. Without the details of the angle we can still find the height the bullet reaches, before hitting the target. Since the bullet starts from the gun at a certain time, it travels both along the x-axis and the y-axis. We use this logic to solve the sum.
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