Answer:
Step-by-step explanation:
This problem involves projectile motion, and we can use the equations of motion to find the various quantities asked for:
i) Equation of path followed by bomb:
The path of the bomb is given by the vector equation r = (50t)i + (4t^2)j, where i and j are unit vectors in the x and y directions, respectively. We can rewrite this equation in terms of x and y coordinates by substituting i = (1, 0) and j = (0, 1), which gives:
x = 50t
y = 4t^2
Therefore, the equation of the path followed by the bomb is y = (1/100)x^2.
ii) Time taken to reach the ground:
The bomb will reach the ground when its height above the ground (y-coordinate) becomes zero. So we can set y = 0 and solve for t:
0 = 4t^2
t = 0 or t = sqrt(0) = 0
This means the bomb will hit the ground at t = 0 and stay on the ground afterwards.
iii) Horizontal distance traversed by the bomb:
The horizontal distance traversed by the bomb is equal to the displacement in the x-direction, which is given by:
Δx = x(final) - x(initial) = 50t - 50(0) = 50t
At t = 0 (when the bomb hits the ground), Δx = 0. Therefore, the bomb travels a horizontal distance of 50t before hitting the ground.
iv) Displacement, velocity and acceleration at t=5sec:
At t=5sec, we can find the displacement of the bomb by substituting t=5 into the vector equation of the path:
r(5) = (50(5))i + (4(5^2))j = 250i + 100j
Therefore, the displacement of the bomb at t=5sec is 250i + 100j.
To find the velocity and acceleration at t=5sec, we can differentiate the vector equation of the path with respect to time:
v = dr/dt = (50)i + (8t)j
a = d^2r/dt^2 = 0i + 8j
Substituting t=5 into these equations, we get:
v(5) = (50)i + (8(5))j = 50i + 40j
a(5) = 0i + 8j
Therefore, the velocity of the bomb at t=5sec is 50i + 40j, and the acceleration is 8j.
v) Tangential and normal component of acceleration at t=5 sec:
The acceleration at t=5sec is 8j. The tangential component of acceleration is zero, since the velocity vector is purely horizontal at this time. The normal component of acceleration is equal to the magnitude of the acceleration vector, since it is perpendicular to the velocity vector. Therefore, the tangential component of acceleration is zero, and the normal component of acceleration is 8 m/s^2.