A body is projected in vertical direction. The altitude y is given by....
Calculation of Initial Velocity of a Projected Body
When a body is projected in the vertical direction, its altitude can be given as y = ut + (1/2)gt^2, where u is the initial velocity of the body, t is the time for which it is projected, and g is the acceleration due to gravity.
Given, y = 10t^2 - 9t + 5
Step 1: Find the Derivative of Altitude with Respect to Time
To determine the initial velocity of the body, we need to find the derivative of altitude with respect to time, which gives us the velocity at any given time.
dy/dt = 20t - 9
Step 2: Determine the Velocity at t = 0
The initial velocity of the body is the velocity at the time of projection, which is t = 0.
Therefore, substituting t = 0 in the above equation, we get
dy/dt = 20(0) - 9 = -9
Therefore, the initial velocity of the body is -9 m/s.
Step 3: Interpretation of the Result
The negative sign indicates that the body was projected in the upward direction, i.e., against the direction of gravity. The magnitude of the initial velocity is 9 m/s, which means that the body was projected with a speed of 9 m/s.
Conclusion
Hence, the initial velocity of the projected body is -9 m/s, and it was projected in the upward direction with a speed of 9 m/s.
A body is projected in vertical direction. The altitude y is given by....
L think -9m/s
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