Question

A ball is projected vertically upwards from top of tower with velocity 50m/s making angle 30^ height of tower 70m after how much time will ball reach ground

Answer 1

Answer:

42.5s

Explanation:

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Answer 2

Dear student,

Given :-

• velocity of projection, v = 50 m/s
• angle of projection, ∅ = 30°
• height above the ground, s = 70 m
• acceleration, a = g = 10 m/s²

To find :-

• the time after which the ball will reach the ground

Solution :-

Let us draw the free body diagram of the given case. Consider the horizontal and vertical components of the projected ball.

  Horizontal component = v cos∅

  Vertical component = v sin∅

From the scenario, it can be observed that only the vertical velocity is effective in this case so we consider the vertical velocity only. So,

Velocity of projection is given by;

     Vp = v sin∅

     Vp = 50 × sin 30°

     Vp = 50 × 1/2

    Vp = 25 m/s

Now, for the time needed, we know by the Newtonian equation of motion that;

 

             S = ut + 1/2 at²

Putting the given values in the above formula, we obtain;

 70 = - 25(t) + 1/2 (10) (t²)

 70 = - 25t + 5t²

 5t² - 25t - 70 = 0

 t² - 5t - 14 = 0

 t² - 7t + 2t - 14 = 0

 t(t - 7) + 2(t- 7) = 0

 (t + 2) =0, (t - 7) =0

 t = - 2, t = + 7

 t = + 7 ( valid )

Hence, the time after which the ball reaches the ground is 7 seconds.