The discussion focuses on calculating the launch velocity and maximum height of a projectile with a time of flight of 41 seconds and a range of 11,750 meters. The key equations involved are h = ut - 1/2gt² for maximum height and the need to determine both vertical and horizontal velocities. The vertical velocity at maximum height is zero, and the horizontal velocity can be derived from the range and time of flight. The correct approach combines these calculations to find the overall launch velocity.
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What is the formula for vertical velocity? What is it at maximum height? That will give you the initial vertical velocity. From that, you can use your formula to find maximum height (Hint: what is t at maximum height?).Awsom Guy said:Homework Statement
A projectile is launched in such a way as to have a time of flight of 41s and a range of 11 750m.
Calculate:
a) the maximum height reached;
b) the launch velocity of the projectile.Homework Equations
h=ut-1/2gt^2The Attempt at a Solution
h=-1/2gt^2
-1/2 * 9.8 * 41^2
I don't get the answer.