Consider a panicle launched at a horizontal velocity v_0 from a height h above t

Question

Consider a panicle launched at a horizontal velocity v_0 from a height h above the ground. Derive an expression for the time it takes the projectile to strike the ground. Ignore air resistance. Derive an expression for the maximum horizontal distance ("range") travelled by the projectile. Ignore air resistance. Let's include the effect of constant air resistance on the projectile: the acceleration of the particle is now given by a = -epsilon gi - gj, where g = 9.8 m/s^2 and epsilon is a very small non-negative constant (epsilon

Explanation / Answer

Given:
vox = initial horizontal velocity = v0
voy = initial vertical velocity = 0
ho = initial height = h

known:
ax = horizontal acceleration = 0
ay = vertical acceleration = -g

1)
y = (voy)(t) + (1/2)(ay)(t^2)

where y = vertical displacment
where voy = initial vertical velocity
where t = time
where ay = vertical acceleration

-h = 0(t) + (1/2)(-g)(t^2)
-h = -0.5gt^2
2h = gt^2
t^2 = 2h/g
t = sqrt[2h/g]

2)
x = (vox)(t) + (1/2)(ax)(t^2)

where x = horizontal displacement
where vox = initial horizontal velocity
where t = time
where ax = horizontal acceleration

x = (v0)(sqrt[2h/g]) + (1/2)(0)(2h/g)
x = v0 * sqrt[2h/g]

cannot answer the bottom questions because some of the constants did not appear

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