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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 )

A ball is thrown up from the top of a tower with an initial velocity of 10 m/s at an angle of 30o with the horizontal. It hits the ground at a distance of 17.3m from the base of tower. Calculate the height of the tower. (g=10m/s2)

Grade:12th pass

2 Answers

Vikas TU
14149 Points
3 years ago
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 / adjacent
Tan 30° = h / 17.3
h = Tan 30 × 17.3
h = 0.5774 × 17.3 = 9.99 M
Vikas Amritiya
askIITians Faculty 230 Points
3 years ago

 

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=53​17.3​=2 secs
 
using equation of motion we get:-
−h=ut−21​gt2
 
=10×2−21​10×22=−10
 
⟹ Height of tower, h=10 m

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