Projectile Problem: Distance and Max Height

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A projectile launched at a 45° angle with an initial velocity of 100√2 ft/sec follows the trajectory described by the equation y = x - (x/25)² for y ≥ 0. To determine how far the projectile travels horizontally before hitting the ground, the horizontal distance can be calculated using the equation's roots. The maximum height is found at the vertex of the parabola, which can be determined by using the vertex formula for parabolas. The projectile's maximum height and horizontal distance are key factors in understanding its motion. Overall, the discussion focuses on calculating the projectile's range and peak height based on its parabolic trajectory.
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Suppose that a projectile is fired at an angle of 45\circ from the
horizontal. Its initial position is the origin in the xy-plane, and its
initial velocity is 100\sqrt{}2 ft / sec. Then its trajectory will be the part of
the parabola y = x -( \frac{}{}x/25)^2 for which y ≥ 0.
a) How far does the projectile travel (horizontally) before it hits the
ground?
b) What is the maximum height above the ground that the
projectile attains?
 
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a) When it hits the ground again, what's the value of y?
b) The highest point is at the vertex of the parabola. Do you know how to find the vertex from the given eqn?
 
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