Question

A ball is projected from the ground into the air. At a height of 15 m, its velocity is observed to be V = 71-10j in m/s. The maximum height the ball will rise is (g = 10 m/s2 options :10 m,,20 m,,,12.5 m,,O15 m​

Answer 1

Here u= 15 m/s, θ=600,g=9.8ms2

Horizontal range,

x=u2sin2θg

= ((15)2 sin (2x60∘)/g 19.88m`

IN the first case the wall is 5 m away from projection point, so it is in the horizontal range of projectile. So the ball will hit he wall. ltbrlt IN the second case (22 m away) wall is notk within the horizontal range. So the ball would not hit the wall.

Answer 2

The maximum height the ball will rise is (b) 20 m.

Given: The velocity of a ball at a height of 15 m is V = 7 î - 10 ĵ in m/s.

To Find: The maximum height the ball will rise.

Solution:

• A vector consists of î and ĵ. î represents the magnitude in the x-direction while ĵ represents the magnitude in the y-direction.

• Whenever the question is concerned about the height or depth in motion, we need to consider the vertical direction or the y-direction (as per the convention).

• So, we can find the initial velocity using the formula of motion which states that,

                v² = u² + 2gH                                                                 ......(1)

Where v = final velocity, u = initial velocity, H = height or depth.

• We know that at the highest position the final velocity of a body becomes zero, so that maximum height can be found using the formula,

                 Hmax = u² / 2g                                                              ......(2)

Where, Hmax = maximum height, u = initial velocity.

Coming to the numerical, it is said that at a height of 15 m, its velocity is observed to be V = 7 î - 10 ĵ in m/s.

So, we can say that,

Vx = 7 m/s and Vy = 10 m/s

First, we need to find the initial velocity in the y-direction, so we use the formula (1),

    Vy² = Uy² + 2gH            

⇒ 10²  = Uy² + 2 × 10 × ( - 15 )   [ As the velocity decreases on going up]

⇒ Uy² = 400

⇒ Uy = √400

⇒ Uy = 20 m/s

Now, to find the maximum height, we know that,

Uy = 20 m/s, Vy = 0, g = 10 m/s².

Putting respective values in (2),

           Hmax = Uy² / 2g    

       ⇒  Hmax = 20² / ( 2 × 10 )

       ⇒  Hmax = 400 / 20

       ⇒  Hmax = 20 m

Hence, the maximum height the ball will rise is (b) 20 m.

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