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The equations of motion of a projectile are given by x = 36t m and 2y = 96t - 9.8t2 m. The angle of projection is
- a)sin−1(4/5)
- b)sin−1(3/5)
- c)sin−1(4/3)
- d)sin−1(3/4)
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The equations of motion of a projectile are given by x = 36t m and 2y ...
Given x = 36t
and 2y = 96t − 9.8t2
or y = 48t − 4.9t2
Let the initial velocity of projectile be u and angle of projection θ. Then, Initial horizontal component of velocity,
and 2y = 96t − 9.8t2
or y = 48t − 4.9t2
Let the initial velocity of projectile be u and angle of projection θ. Then, Initial horizontal component of velocity,
Most Upvoted Answer
The equations of motion of a projectile are given by x = 36t m and 2y ...
The equation of motion for the x-direction is given by x = 36t.
The equation of motion for the y-direction is given by 2y = 96t - 9.8t^2.
We can rearrange the equation of motion for the x-direction to solve for t:
t = x/36.
Substituting this value of t into the equation of motion for the y-direction, we get:
2y = 96(x/36) - 9.8(x/36)^2.
Simplifying this equation, we get:
2y = 8x - 0.27x^2.
Now, let's find the maximum height of the projectile. To do this, we set the y-coordinate to 0 and solve for x:
2(0) = 8x - 0.27x^2.
0 = x(8 - 0.27x).
x = 0 or x = 29.63.
Since x represents time, we discard the solution x = 0, as it does not make physical sense.
Now, let's find the time at which the projectile hits the ground. To do this, we set the y-coordinate to 0 and solve for t:
2(0) = 96t - 9.8t^2.
0 = t(96 - 9.8t).
t = 0 or t = 9.795.
Since t represents time, we discard the solution t = 0, as it does not make physical sense.
Now, let's find the angle of projection. We can use the equation tan(theta) = y/x to find the angle:
tan(theta) = y/x = (96t - 9.8t^2)/(36t).
Simplifying this equation, we get:
tan(theta) = (96 - 9.8t)/(36).
Substituting the value of t = 9.795 into this equation, we get:
tan(theta) = (96 - 9.8(9.795))/(36) = 0.82.
The angle of projection can be found by taking the inverse tangent of both sides:
theta = arctan(0.82) = 39.1 degrees.
Therefore, the angle of projection is approximately 39.1 degrees.
The equation of motion for the y-direction is given by 2y = 96t - 9.8t^2.
We can rearrange the equation of motion for the x-direction to solve for t:
t = x/36.
Substituting this value of t into the equation of motion for the y-direction, we get:
2y = 96(x/36) - 9.8(x/36)^2.
Simplifying this equation, we get:
2y = 8x - 0.27x^2.
Now, let's find the maximum height of the projectile. To do this, we set the y-coordinate to 0 and solve for x:
2(0) = 8x - 0.27x^2.
0 = x(8 - 0.27x).
x = 0 or x = 29.63.
Since x represents time, we discard the solution x = 0, as it does not make physical sense.
Now, let's find the time at which the projectile hits the ground. To do this, we set the y-coordinate to 0 and solve for t:
2(0) = 96t - 9.8t^2.
0 = t(96 - 9.8t).
t = 0 or t = 9.795.
Since t represents time, we discard the solution t = 0, as it does not make physical sense.
Now, let's find the angle of projection. We can use the equation tan(theta) = y/x to find the angle:
tan(theta) = y/x = (96t - 9.8t^2)/(36t).
Simplifying this equation, we get:
tan(theta) = (96 - 9.8t)/(36).
Substituting the value of t = 9.795 into this equation, we get:
tan(theta) = (96 - 9.8(9.795))/(36) = 0.82.
The angle of projection can be found by taking the inverse tangent of both sides:
theta = arctan(0.82) = 39.1 degrees.
Therefore, the angle of projection is approximately 39.1 degrees.
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The equations of motion of a projectile are given by x = 36t m and 2y = 96t - 9.8t2 m. The angle of projection isa)sin−1(4/5)b)sin−1(3/5)c)sin−1(4/3)d)sin−1(3/4)Correct answer is option 'A'. Can you explain this answer?
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