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Answer :
Final answer:
To find the initial velocity of a ball thrown upwards, analyze its motion past a window and the time it takes. The correct answer is B.
Explanation:
To find the initial velocity of the ball thrown upwards, we can use the concept of projectile motion. The ball passes a window 1.70m high at a point 7.50m above the ground, and it takes 1.15 seconds to go past the window. We can use the equation for vertical motion:
[tex]\[ h = v_{0}t - \frac{1}{2}gt^{2} \][/tex]
where:
- h is the displacement (1.70m),
- [tex]\( v_{0} \)[/tex] is the initial velocity (which we're solving for),
- t is the time (1.15s), and
- g is the acceleration due to gravity [tex](-9.8 m/s^2)[/tex].
Rearranging the equation to solve for [tex]\( v_{0} \)[/tex], we get:
[tex]\[ v_{0} = \frac{h + \frac{1}{2}gt^{2}}{t} \][/tex]
Plugging in the values, we find [tex]\( v_{0} = \frac{1.70m + \frac{1}{2}(-9.8m/s^2)(1.15s)^2}{1.15s} \approx 5.2 \, \text{m/s} \)[/tex]. Therefore, the correct answer is B) 5.2 m/s.
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Rewritten by : Jeany
Final Answer:
The initial velocity of the ball when it was thrown upward was 5.2 m/s. The correct answer is option B) 5.2 m/s.
Explanation:
To find the initial velocity of the ball, we can use the equations of motion for uniformly accelerated motion. Since the ball passes a window 1.70 m above the ground, we know that its maximum height reached is 1.70 m. Using the equation for maximum height in vertical motion, hₘâ‚â‚“ = (váµ¢â¿áµ¢â‚œáµ¢â‚â‚—²) / (2 * g), where váµ¢â¿áµ¢â‚œáµ¢â‚â‚— is the initial velocity and g is the acceleration due to gravity (9.8 m/s²), we can solve for váµ¢â¿áµ¢â‚œáµ¢â‚â‚—.
Rearranging the equation gives us váµ¢â¿áµ¢â‚œáµ¢â‚â‚— = √(2 * g * hₘâ‚â‚“). Substituting the given values, we find váµ¢â¿áµ¢â‚œáµ¢â‚â‚— = √(2 * 9.8 * 1.70) ≈ 5.2 m/s.
Therefore, the initial velocity of the ball when it was thrown upward was 5.2 m/s, corresponding to option B) in the question.
