Answer
Verified
435k+ views
Hint: Here initial velocity, distance from the target and gravity are given. We need to find the position at which centre of the target rifle must be aimed in order to hit the target. When the bullet is fired, it will have a projectile motion. We have learned the relation between distance covered, velocity and gravity of projectile motion. By using this horizontal range equation, we can solve the question.
Formula used:
\[h=ut+\dfrac{1}{2}g{{t}^{2}}\]
\[velocity, v=\dfrac{displacement}{time}\]
Complete answer:
Given,
\[\text{Initial velocity of bullet = 630 m/s}\]
\[\text{Distance from the center of target = 700 m}\]
\[g=10m/{{s}^{2}}\]
Let’s find the solution using vertical height equation,
\[h=ut+\dfrac{1}{2}g{{t}^{2}}\]------ 1
Where,
\[\text{u - initial velocity}\]
\[\text{h - vertical height}\]
\[\text{g- acceleration due to gravity}\]
\[\text{t- time taken for the vertical fight}\]
For vertical motion of the bullet,
\[u=0\]------- 2
Let, \[t\] be the time taken by the bullet to hit the target with velocity \[v=630m/s\], and displacement, \[d=700m\]. Then,
\[time=\dfrac{displacement}{velocity}=\dfrac{700}{630}=\dfrac{10}{9}s\] -------- 3
Substitute 2 and 3 in equation 1. We get,
\[h=\dfrac{1}{2}\times 10\times {{\left( \dfrac{10}{9} \right)}^{2}}=\dfrac{500}{81}=6.1m\]
Therefore, the bullet should be fired \[6.17m\] above the center of target.
So, the correct answer is “Option D”.
Additional Information:
According to Newton's first law of motion a bullet fired from a gun will continue in motion at the same speed unless an external force is acted on it. The horizontal motion of the bullet is constant. Vertically, the bullet is uniformly accelerated by gravity. Ignoring air resistance, the only force acting on a bullet is gravity. The downward motion of a bullet is the same as a body in free-fall. Hence bullets fired from a rifle have a projectile path.
Note:
We have,
\[h=ut+\dfrac{1}{2}g{{t}^{2}}\]
For a vertical motion of the bullet,
\[u=0\]
Then,
\[t=\sqrt{\dfrac{2h}{g}}\]---- a
Given,
\[h=700m\]
\[g=10m/{{s}^{2}}\]
We know,
\[velocity, v=\dfrac{displacement}{time}=\dfrac{d}{t}\]
Then,
\[d=vt\]--- b
Substitute equation a in b. We get,
\[d=\sqrt{\dfrac{2h}{g}}\times v\] ---- c
Substituting the values of \[\text{d,g and v}\]in equation c, we get,
\[700=\sqrt{\dfrac{2\times h}{0}}\times 630\]
Then,
\[h=6.17m\]
Formula used:
\[h=ut+\dfrac{1}{2}g{{t}^{2}}\]
\[velocity, v=\dfrac{displacement}{time}\]
Complete answer:
Given,
\[\text{Initial velocity of bullet = 630 m/s}\]
\[\text{Distance from the center of target = 700 m}\]
\[g=10m/{{s}^{2}}\]
Let’s find the solution using vertical height equation,
\[h=ut+\dfrac{1}{2}g{{t}^{2}}\]------ 1
Where,
\[\text{u - initial velocity}\]
\[\text{h - vertical height}\]
\[\text{g- acceleration due to gravity}\]
\[\text{t- time taken for the vertical fight}\]
For vertical motion of the bullet,
\[u=0\]------- 2
Let, \[t\] be the time taken by the bullet to hit the target with velocity \[v=630m/s\], and displacement, \[d=700m\]. Then,
\[time=\dfrac{displacement}{velocity}=\dfrac{700}{630}=\dfrac{10}{9}s\] -------- 3
Substitute 2 and 3 in equation 1. We get,
\[h=\dfrac{1}{2}\times 10\times {{\left( \dfrac{10}{9} \right)}^{2}}=\dfrac{500}{81}=6.1m\]
Therefore, the bullet should be fired \[6.17m\] above the center of target.
So, the correct answer is “Option D”.
Additional Information:
According to Newton's first law of motion a bullet fired from a gun will continue in motion at the same speed unless an external force is acted on it. The horizontal motion of the bullet is constant. Vertically, the bullet is uniformly accelerated by gravity. Ignoring air resistance, the only force acting on a bullet is gravity. The downward motion of a bullet is the same as a body in free-fall. Hence bullets fired from a rifle have a projectile path.
Note:
We have,
\[h=ut+\dfrac{1}{2}g{{t}^{2}}\]
For a vertical motion of the bullet,
\[u=0\]
Then,
\[t=\sqrt{\dfrac{2h}{g}}\]---- a
Given,
\[h=700m\]
\[g=10m/{{s}^{2}}\]
We know,
\[velocity, v=\dfrac{displacement}{time}=\dfrac{d}{t}\]
Then,
\[d=vt\]--- b
Substitute equation a in b. We get,
\[d=\sqrt{\dfrac{2h}{g}}\times v\] ---- c
Substituting the values of \[\text{d,g and v}\]in equation c, we get,
\[700=\sqrt{\dfrac{2\times h}{0}}\times 630\]
Then,
\[h=6.17m\]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The polyarch xylem is found in case of a Monocot leaf class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Casparian strips are present in of the root A Epiblema class 12 biology CBSE
How do you graph the function fx 4x class 9 maths CBSE