A body is launched upwards at 60 m/s at 30 degrees to the horizontal. What is the maximum height of the body and maximum distance at which it falls.

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The body is launched upwards with an initial velocity 60 m/s and the angle of projection with the horizontal is 30 degrees. To determine the maximum distance traveled by the body in the horizontal and vertical directions, we take the vertical and horizontal components of the velocity. Its vertical component is 60*sin 30 = 60*(1/2) = 30 m/s and the horizontal component is `60*(sqrt 3/2)` = `30*sqrt 3` m/s. As the object moves, its horizontal component of velocity decreases at 9.8 m/s^2 due to the gravitational acceleration in the downwards direction. The horizontal component of its velocity remains unchanged.

At the highest point of its path, the vertical component is equal to 0. The height of the body at this point is given by the equation v^2 - u^2 = 2*g*s where s is the height.

Substituting the values for the variable gives:

0 - 30^2 = 2*9.8*s

s = 900/19.6 = 45.918 m

The time taken by the body to reach the highest point is equal to t, given by the equation v = u + g*t

t = 30/9.8 = 3.06

It returns to the surface after a time 2*t = 6.12 s

The horizontal distance traveled in 6.12 s is s = u*t + (1/2)*a*t^2. Substituting the values for the variables gives s = `30*sqrt 3*6.12` = 318.13

The body has a maximum height of 45.918 m and the horizontal distance traveled before it comes back to the surface is 318.13 m

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