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ArbitrLikvidat [17]
3 years ago
8

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t

he velocity at launch?
Physics
1 answer:
Lyrx [107]3 years ago
6 0

Given that

Velocity of missile (v) = 20 m/s ,

Angle of missile (Θ) = 53°

Determine ,   Vertical component  = v sin Θ

                                                        = 20 sin 53°

                                                        = 15.97 m/s

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Answer:

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Why doesn’t the ball roll on forever after being kicked at a soccer game?
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My guess would be because the gravity from the Earth's core is constantly pulling the ball towards the ground. It's like the moon. Why doesn't the moon just float away in space? Because Earth's gravitational pull keeps it rotating around it. Therefore, the ball will always be pulled towards the core which keeps it from from rolling forever due to friction. But i may be wrong, even though this a quite a good answer, hope it is right!

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4 years ago
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
algol13

Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
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The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

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