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luda_lava [24]
3 years ago
7

A van is traveling on a road at a speed of 55 km/h. A girl sitting near the driver of the van throws a paper airplane to a boy a

t the back of the van with a speed of 2 km/h.
The speed of the paper airplane, to the nearest whole number relative to a stationary observer on the side of the road, is ___ km/h.
Physics
2 answers:
e-lub [12.9K]3 years ago
8 0

Answer:

53 km/h

Explanation:

Let's take as positive direction the direction in which the van is moving. In this case, we can write as follows:

- Velocity of the van relative to the road: v_v = 55 km/h

- Velocity of the paper airplane relative to the van: v_a = -2 km/h (with a negative sign since it is moving into opposite direction)

Therefore, the velocity of the paper airplane relative to the road (stationary observer) will be

v=v_v + v_a = 55 km/h - 2 km/h = 53 km/h

aivan3 [116]3 years ago
5 0
<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.

The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h.  No rounding is necessary.</span>
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a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^
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Approximately 0.8\; \rm s (assuming that air resistance is negligible.)

Explanation:

Let v_0 denote the initial velocity of this ball. Let \theta denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

  • Initial vertical velocity: v_0(\text{vertical}) = v_0 \cdot \sin(\theta).
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If air resistance on this ball is negligible, v_0(\text{vertical}) alone would be sufficient for finding the time of flight of this ball.

Calculate v_0(\text{vertical}) given that v_0 = 8 \; \rm m \cdot s^{-1} and \theta = 30^\circ:

\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}.

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}, the vertical velocity of this ball right before landing would be v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}.

Calculate the change to the vertical velocity of this ball:

\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}.

In other words, the vertical velocity of this ball should have change by 8\; \rm m \cdot s^{-1} during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is g = 10\; \rm m \cdot s^{-2}. In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by 10\; \rm m\cdot s^{-1} each second. What fraction of a second would it take to change the vertical velocity of this ball by 8\; \rm m \cdot s^{-1}?

\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}.

In other words, it would take 0.8\; \rm s to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be 0.8\; \rm s\!.

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