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velikii [3]
3 years ago
5

A cricketer can throw a ball to a maximum horizontal distance of 100m.How much high above the ground can he throw same ball

Physics
1 answer:
rosijanka [135]3 years ago
5 0
<span>The range of an object thrown at a velocity v with an angle of elevation 45 degrees is S = v^2/g 100 = v^2/g => v^2 = 100*g In the same throw, the maximum height of the ball can be derived from the equation v^2 - u^2 = 2*g*h It is is h = = = 50 metres</span>
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As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is   v =0.333 \  m/s in positive x -direction

The speed is s = 0.733 \ m/s

Explanation:

From the question we are told that

The distance from the house to truck is  D =  20 m

  The distance traveled back to retrieve  wind-blown hat is  d =  15

  The distance from the wind-blown hat position too the truck is  k =  20  m

  The total time taken is  t  =  75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

      L  =  20 - 15 + 20

=>    L  =  25 \ m

Generally the average velocity is mathematically represented as

          v  =  \frac{L}{t}

=>      v = \frac{25}{75}

=>      v =0.333 \  m/s

Generally the distance covered by Justin is mathematically represented as  

         R =  D+ d + k

=>      R =  20 + 15 +20

=>     R =  55 \  m

Generally Justin's average speed over a 75 s period is mathematically represented as

            s = \frac{R}{ t}

=>         s = \frac{55}{ 75}

=>        s = 0.733 \ m/s

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