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A ball is thrown up from the top of a tower with an initial velocity of 1 0 m/s at an angle of 3 0 o with the horizontal. It hits the ground at a distance of 1 7 . 3 m from the base of tower. Calculate the height of the tower. ( g = 1 0 m / s 2 )
Dear student When we draw the diagram of the motion we get a right angled triangle.We have an angle of 30° and a base of 17.3M.We will use trigonometric ratios to solve this.We will use tan.Tan = Opposite / adjacentTan 30° = h / 17.3h = Tan 30 × 17.3h = 0.5774 × 17.3 = 9.99 M
The ball is thrown at an angle, θ=30o.Initial velocity of the ball, u=10 m/s Horizontal range of the ball, R=17.3 m We know that, R=u cosθ t, where t is the time of flight ⟹t=5317.3=2 secs using equation of motion we get:-−h=ut−21gt2 =10×2−2110×22=−10 ⟹ Height of tower, h=10 m
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