The path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet,where the model for the path is $$x ^ { 2 } = - \frac { v ^ { 2 } } { 16 } ( y - s )$$ .In this model (in which air resistance is disregarded),y is the height (in feet)of the projectile and x is the horizontal distance (in feet)the projectile travels.A ball is thrown from the top of a 150-foot tower with a velocity of 32 feet per second.Find the equation of the parabolic path.
A) $x ^ { 2 } = - 32 ( y - 150 )$
B) $x ^ { 2 } = 64 ( y - 150 )$
C) $x ^ { 2 } = 1406 ( y - 32 )$
D) $x ^ { 2 } = - 1406 ( y - 32 )$
E) $x ^ { 2 } = - 64 ( y - 150 )$

Question

The path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet,where the model for the path is $$x ^ { 2 } = - \frac { v ^ { 2 } } { 16 } ( y - s )$$ .In this model (in which air resistance is disregarded),y is the height (in feet)of the projectile and x is the horizontal distance (in feet)the projectile travels.A ball is thrown from the top of a 150-foot tower with a velocity of 32 feet per second.Find the equation of the parabolic path. ​
A) $x ^ { 2 } = - 32 ( y - 150 )$
B) $x ^ { 2 } = 64 ( y - 150 )$
C) $x ^ { 2 } = 1406 ( y - 32 )$
D) $x ^ { 2 } = - 1406 ( y - 32 )$
E) $x ^ { 2 } = - 64 ( y - 150 )$

Quizplus · Accepted Answer

Substituting $v = 32$ feet per second and $s = 150$ feet into the given model, we get $x^2 = -\frac{32^2}{16}(y - 150)$, which simplifies to $x^2 = -64(y - 150)$.

The Path of a Projectile Projected Horizontally with a Velocity x2=−v216(y−s)x ^ { 2 } = - \frac { v ^ { 2 } } { 16 } ( y - s )x2=−16v2​(y−s)

The Path of a Projectile Projected Horizontally with a Velocity $x ^ { 2 } = - \frac { v ^ { 2 } } { 16 } ( y - s )$