A body is projected at an angle of 30degree to the horizontal with a velocity of 150metre per seconds.Calculate the time it takes to reach the greatest height,take g=10metre per seconds square and neglect air resistance.
v = Vi - g t
here Vi = 150 sin 30 = 75 m/s
at top v = 0
0 = 75 - 10 t
solve for t at top
t = 7.5 seconds
then if they ask for the height
h = (1/2) g t^2 to fall from top
= 5 * (7.5)^2
then if they ask for range
u = 150 cos 30
range = u t = 150 cos 30 * 7.5 * 2
the 2 is because t is only the upward half of the parabola
Another answer
Thanks it is very explanatory
To calculate the time it takes for the body to reach the greatest height, we need to break down the initial velocity into its horizontal and vertical components.
Given:
Angle of projection (θ) = 30 degrees
Initial velocity (u) = 150 m/s
Horizontal component (u_x) can be calculated as:
u_x = u * cos(θ)
Vertical component (u_y) can be calculated as:
u_y = u * sin(θ)
Let's calculate the horizontal and vertical components of the initial velocity:
u_x = 150 m/s * cos(30 degrees)
u_y = 150 m/s * sin(30 degrees)
u_x = 150 m/s * 0.866
u_y = 150 m/s * 0.5
u_x = 129.9 m/s (approx.)
u_y = 75 m/s (approx.)
Now, we can calculate the time taken to reach the greatest height using the equation for vertical motion:
u_y = u * sin(θ) = u_initial * sin(θ) - g * t
Since the body reaches its greatest height when its vertical velocity becomes zero, we can set u_y = 0 and solve for t:
0 = u_y - g * t
Rearranging the equation:
t = u_y / g
Substituting the values:
t = 75 m/s / 10 m/s^2
t = 7.5 s
Therefore, it will take approximately 7.5 seconds for the body to reach the greatest height.