Answer:
maximum height is y = b²/18g √ (12L/b)³
Explanation:
Let's analyze the situation first we have a projectile subjected to an acceleration depends on time, so we must use the definition of acceleration to find the speed when it is at distance L, then we will use the projectile launch equations
Acceleration dependent on t
a = dv / dt
dv = adt
∫dv =∫ (b t²) dt
v = b t³ / 3
The initial speed is zero for zero time
we use the definition of speed
v = dy / dt
dy = v dt
∫dy = ∫b t³ / 3 dt
y = b/3 t⁴ / 4
y = b/12 t⁴
we evaluate from the initial point where the height is zero for the zero time
Let's calculate the time to travel the length (y = L) of the canyon
t = (12 y / b) ¼
t = (12 L / b) ¼
Taking the time, we can calculate the projectile's output speed
v = b/3
This is the speed of the body, which is the initial speed for the projectile launch movement. Let's calculate the highest point where the zero speed
Vy² = v₀² - 2 g y
0 = Vo² - 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