Question

Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us assume that the projectile lands on a spot that is at a height h = 10 feet, raised in elevation from its launch point. For this entire problem, you can neglect drag.
a) Determine an expression for the horizontal range R as a function of time. Here, range represents the total horizontal distance traveled from launch point to point of landing. Your answer will be in terms of initial velocity v0, time t, height h and acceleration due to gravity, g. At this point, do not substitute any of the numerical values yet.
b) Determine the time taken to achieve the maximum possible range, R. Think about maxima and minima from calculus when solving this part of the problem.
c) Once you have determined the time from part (b), substitute the numerical values for the terms and obtain the actual time in seconds. Then, determine the initial launch angle, θ (also called the angle of attack), such that the projectile achieves the maximum possible horizontal range.
d) What is the numerical value of this maximum R (based on your calculations from parts (b) and (d)).
e) Determine the velocity at the peak of the trajectory of the projectile.
f) Determine the time taken for the projectile to reach its peak height on its trajectory.
g) Determine the peak height reached by the projectile.

Answer 1

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation: