Question

In a Fourth of July celebration, a firecracker is launched from ground level with an initial velocity of 17.0 m/s at 28.8 ∘ from the vertical. At its maximum height it explodes in a starburst into many fragments, two of which travel forward initially at 28.4 m/s at ± 40.3 ∘ with respect to the horizontal, both quantities measured relative to the original firecracker just before it exploded.

With what angles with respect to the horizontal do the two fragments initially move right after the explosion, as measured by a spectator standing on the ground?

Answer 1

At the maximum height, the firecracker had only the horizontal velocity.
v = vo sin
= 17.0 x sin(28.8)
= 8.19 m/s

Horizontal velocity of the fragment with respect to the firecracker,
v_xi = 28.4 x cos(40.3)
= 21.66 m/s
Horizontal velocity of the fragment with respect to the spectator,
v_x = v_xi + v
= 21.66 + 8.19
= 29.85 m/s

Vertical velocity of the fragment remains unchanged.
v_y = 28.4 sin(40.3)
= 18.37

Angle measured by spectator = tan^-1(v_y/v_x)
= tan^-1(18.37/29.85)
= 31.61 degrees.