A body is thrown at an angle 30 to the horizontal with the velocity of 30m/s.After 1 second,its velocity will be
To find the velocity of the body after 1 second, we can break down the initial velocity into its horizontal and vertical components. The horizontal component can be found using the formula:
Vx = V * cos(theta)
where:
Vx is the horizontal component of velocity
V is the initial velocity (30 m/s in this case)
theta is the angle with respect to the horizontal (30° in this case)
Vx = 30 m/s * cos(30°)
Vx = 30 m/s * √3/2
Vx = 30 * √3 / 2
Vx ≈ 15√3 m/s
The vertical component can be found using the formula:
Vy = V * sin(theta)
where:
Vy is the vertical component of velocity
V is the initial velocity (30 m/s in this case)
theta is the angle with respect to the horizontal (30° in this case)
Vy = 30 m/s * sin(30°)
Vy = 30 m/s * 1/2
Vy = 15 m/s
Now, after 1 second, we need to consider the effects of gravity on the vertical component. The velocity in the vertical direction changes due to the acceleration due to gravity, which is approximately 9.8 m/s². So we can calculate the vertical velocity after 1 second using the formula:
Vy = Vy(initial) + (acceleration due to gravity) * time
Vy(after 1 second) = Vy - 9.8 m/s² * 1s
Vy(after 1 second) = 15 m/s - 9.8 m/s² * 1
Vy(after 1 second) ≈ 15 m/s - 9.8 m/s
Vy(after 1 second) ≈ 5.2 m/s
Therefore, after 1 second, the velocity of the body will be approximately 15√3 m/s horizontally and 5.2 m/s vertically.
v = velocity up = 30 sin 30 - 9.81 t
u = velocity horizontal = 30 cos t
if t = 1
v = 15 - 9.81
u = 30(.866)
speed = sqrt (u^2+v^2)
tan angle up = v/u