A projectile of mass M, initially at rest, is acted upon by a net force [including gravity] that increases quadratically with ti

Answer:

The peak height can be expressed as y = b²/18g √(12L/b)³

Explanation:

We begin by assessing the scenario involving a projectile subject to time-dependent acceleration. Thus, we must define acceleration to determine the speed when at distance L, then apply the equations governing projectile launch.

Acceleration that varies with time:

a = dv / dt

dv = a dt

∫dv = ∫(bt²) dt

v = (b t³)/3

The starting speed is zero at t = 0.

Utilizing the definition of speed:

v = dy / dt

dy = v dt

∫dy = ∫(b t³/3) dt

y = (b/3)(t⁴/4)

y = (b/12)t⁴

From the initial instance where height is zero at time zero, we now compute the time needed to traverse the distance (y = L) across the canyon:

t = (12y/b)¼

t = (12L/b)¼

This time allows us to find the projectile's launch speed:

v = (b/3)

This speed acts as the initial speed for the projectile movement. To find its maximum height at zero final speed:

Vy² = v₀² - 2 g y

0 = v₀² - 2 g y

2 g y = v₀²

y = v₀²/2g

y = (1/2g)([b/3 ])²

y = (1/2g)(b²/9 )

y = b²/18g √(12L/b)³