If a body is projected upward with a velocity of 40 root 2 metre per s...
Problem Statement:
A body is projected upward with a velocity of 40 root 2 metre per second at an angle of 45 degree with horizontal under gravity of 10 metre per second from the ground then calculate the velocity of a body at t = 2seconds ,angle made by the velocity with horizontal at t=1 second third angle between v and a at t= 3 second and time equals to 5 seconds?
Solution:
Given: Initial velocity (u) = 40 root 2 m/s, angle of projection (theta) = 45 degrees, acceleration due to gravity (g) = 10 m/s^2
Velocity at t = 2 seconds:
Using the formula to find the velocity of a body at any time t:
v = u + at
Here, u = 40 root 2 m/s, a = g = 10 m/s^2 and t = 2 s
So, v = 40 root 2 + (10 x 2) = 60.14 m/s
Angle made by velocity with horizontal at t = 1 second:
At any time t, the horizontal component of velocity (u cos theta) remains constant, while the vertical component of velocity (u sin theta) changes with time due to the acceleration due to gravity.
At t = 1 s, the vertical displacement of the body is given by:
s = ut sin theta + (1/2)at^2
Here, u = 40 root 2 m/s, theta = 45 degrees, a = g = 10 m/s^2 and t = 1 s
So, s = (40 root 2 x 1/2) + (1/2 x 10 x 1) = 25.7 m
Using the formula to find the angle made by velocity with horizontal at any time t:
tan alpha = (v sin theta)/ (u cos theta - gt)
Here, v is the velocity at time t, u and theta are the initial velocity and angle of projection, and g is the acceleration due to gravity.
At t = 1 s, u = 40 root 2 m/s, theta = 45 degrees, g = 10 m/s^2, v = u - gt (since the body is moving upward)
So, v = 40 root 2 - 10 x 1 = 28.28 m/s
Substituting the values in the formula:
tan alpha = (28.28 sin 45)/ (40 root 2 cos 45 - 10 x 1) = 0.5
So, alpha = tan^-1(0.5) = 26.6 degrees
Angle between velocity and acceleration at t = 3 seconds:
At any time t, the acceleration due to gravity acts vertically downwards, while the velocity of the body has horizontal and vertical components.