Physics, 19.11.2020 15:20 dw685978

A stone is thrown horizontally. In 0.5s

after the stone began to move, the numerical value of its

velocity was 1.5 times its initial velocity. Find the initial velocity of the stone. Disregard the

resistance of the air.

Answers: 3

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Answer from: hawleyemily

Approximately $4.39 \; \rm m \cdot s^{-1}$ (assuming that $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let the initial velocity of this stone be $v\; \rm m \cdot s^{-1}$ ( $v \ge 0$ because speed should be non-negative.) That should be numerically equal to the initial horizontal speed of this stone. Because the stone was thrown horizontally, its vertical speed would initially be zero ( $0\; \rm m \cdot s^{-1}$ .)

The weight of this stone will accelerate the stone downwards. Because the weight of the stone is in the vertical direction, it will have no effect on the horizontal speed of the stone.

Therefore, if air resistance is indeed negligible, the horizontal speed of this stone will stay the same. The horizontal speed of the stone after $0.5\; \rm s$ should still be $v\; \rm m \cdot s^{-1}$ - same as its initial value.

On the other hand, if $g = 9.81\; \rm m \cdot s^{-2}$ , the (downward) vertical speed of this stone will increase by $9.81\; \rm m \cdot s^{-2}$ every second. After $0.5\; \rm s$ , the vertical speed of this stone would have become:

$0.5\; \rm s \times 9.81\; \rm m \cdot s^{-2} = 4.905\; \rm m \cdot s^{-1}$ .

$t = 0\; \rm s$ :

Horizontal speed: $v\; \rm m \cdot s^{-1}$ .Vertical speed: $0\; \rm m \cdot s^{-1}$ .

$t = 0.5\; \rm s$ :

Horizontal speed: $v\; \rm m \cdot s^{-1}$ .Vertical speed: $4.905\; \rm m \cdot s^{-1}$ .

Refer to the diagram attached. Consider the vertical speed and the horizontal speed of this rock as lengths of the two legs of a right triangle. The numerical value of the velocity of this stone would correspond to the length of the hypotenuse of that right triangle. Apply the Pythagorean Theorem to find that numerical value:

$\begin{aligned} &\text{numerical value of velocity} \\ &= \sqrt{{\left(\text{horizontal speed}\right)}^2 + {\left(\text{vertical speed}\right)}^2}\end{aligned}$ .

At $t = 0\; \rm s$ , the magnitude of the velocity of this stone would be $\sqrt{v^{2} + 0^2} = v\; \rm m \cdot s^{-1}$ .

At $t = 0.5\; \rm s$ , the magnitude of the velocity of this stone would be $\left(\sqrt{v^2 + 4.905^2}\right)\; \rm m \cdot s^{-1}$ .

The question requires that the magnitude of the velocity of the stone at $t = 0.5\; \rm s$ should be 1.5 times the value at $t = 0\; \rm s$ . In other words:

$\sqrt{v^2 + 4.905^2} = 1.5 \, v$ .

Square both sides of this equation and solve for :

$v^2 + 4.905^2 = 2.25\, v^2$ .

$v \approx 4.39$ ( $v \ge 0$ given that speed should be non-negative.)

Therefore, the initial velocity of this stone should be approximately $4.39\; \rm m \cdot s^{-1}$ .

A stone is thrown horizontally. In

0.5safter the stone began to move, the numerical value of its ve

Answer from: Quest

not too sure thinking all accelerations are pointed down

Answer from: Quest

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