When a mortar shell is ﬁred with an initial

1 answer:

5 0

The equation y = Vo cos(alfa) t - 16t^2 implies that the mortar landed at the same level that it was fire and that fire angle is also 75°.

With that said, let us work on the parametric equations

y = 0 = Vo sin(alfa) t - 16t^2, which by factoring =>

t (Vo sin(alfa) - 16t) = 0 => t = Vo sin(alfa) / 16 .....(1)

x = 2500 = Vo cos(alfa) t => t = 2500 / [Vo cos(alfa)] ..... (2)

Now make (1) equal to (2)

Vo sin(alfa) / 16 = 2500 / [Vo cos(alfa)] =>

Vo^2 sin(alfa)cos(alfa) = 2500ft*16ft/s^2 =>

Vo^2 = 2500*16 / [sin(75°)cos(75)] = 2500*16/0.25 = 160,000 ft^2/s^2

Vo = √(160,000) ft/s = 400 ft/s

Answer: option 5. 400 ft/s

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g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl

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Answer:

$v=1.5m/s$

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ . Since $v=r\omega$ (the ball rolls without slipping) and for a solid sphere $I=\frac{2mr^2}{5}$ , we have: