Projectile Problem: Distance and Max Height

[GusTu2007]

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?

 

[Shooting Star]

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?

 

[ogb p]

a) To find the distance traveled horizontally, we can use the formula for horizontal range: R = (v^2*sin2θ)/g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. Plugging in the given values, we get R = (100√2)^2*sin2(45)/32.2 = 1000 ft. Therefore, the projectile travels 1000 ft horizontally before hitting the ground.

b) To find the maximum height, we can use the formula for maximum height: H = (v^2*sin^2θ)/2g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. Plugging in the given values, we get H = (100√2)^2*sin^2(45)/64.4 = 1000 ft. Therefore, the projectile reaches a maximum height of 1000 ft above the ground.