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3 years ago

5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)

Physics

1 answer:

jolli1 [7]3 years ago

3 0

Answer:

(B) 13.9 m

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

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4 years ago

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DIA [1.3K]

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3 years ago

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3 years ago

3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed rel

kramer

Answer:

$v_s=\frac{v}{1+\frac{m}{M} }$

Explanation:

Let the shells speed with respect to ground be $v_s$

Let the shells speed with respect to ground be $v_g$

mass of shell, $m$

mass of gun, $M$

The relative velocity of shell with respect to a free unconstrained gun, $v$

According to the Newton's third law of motion the direction of the velocity of gun and shell will be in the opposite direction.

So, the relation between the relative velocity and their individual velocity will be:

$v=v_s+v_g$ ......................(1)

<u>And according to the conservation of momentum (as the condition is very close to the elastic collision):</u>

$M.v_g-m.v_s=0$

substitute the value of $v_g$ from equation (1)

$M\times (v-v_s)=m\times v_s$

$M.v-M.v_s=m.v_s$

$M.v=M.v_s+m.v_s$

$v_s=\frac{M.v}{(M+m)}$

$v_s=\frac{v}{(\frac{(M+m)}{M}) }$

$v_s=\frac{v}{1+\frac{m}{M} }$

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3 years ago