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Tamiku [17]
3 years ago
5

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^

2)​
Physics
1 answer:
posledela3 years ago
5 0

Answer:

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).
  • Initial horizontal velocity: v_0(\text{vertical}) = v_0 \cdot \cos(\theta).

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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