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a ball is thrown with 25 m//s at an angl...

a ball is thrown with `25 m//s` at an angle `53^(@)` above the horizontal. Find its time to fight, maximum height and range.

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AI Generated Solution

To solve the problem of a ball thrown with an initial velocity of \(25 \, \text{m/s}\) at an angle of \(53^\circ\) above the horizontal, we will calculate the time of flight, maximum height, and range step by step. ### Step 1: Calculate the Time of Flight The formula for the time of flight (\(T\)) of a projectile is given by: \[ T = \frac{2u \sin \theta}{g} ...
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A ball is thrown at a speed of 40 m/s at an angle of 60^0 with the horizontal. Find a. the maximum height reached and b. the range of te ball. Take g=10 m/s^2 .

A particle is thrown with speed of 50 m//s at an angle of projection 37^(@) with the horizontal. Find (a) the time of flight, (b) the maximum height attained and ( c) the horizontal range. (sin 37^(@) = (3)/(5), cos 37^(@) = (4)/(5))

Knowledge Check

  • A ball is thrown with speed 40 m//s at an angle 30^(@) with horizontally from the top of a tower of height 60 m . Choose the correct option

    A
    the vertical component of velocity first decreases to zero and then increases
    B
    the ball reaches the ground after `6 s`
    C
    if the ball strikes the ground at maximum horizontal distance from the tower for this the angle of projection should be less than `45^(@)`
    D
    all option are correct

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