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Grade: 12th pass

                        

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 )

2 months ago

Answers : (2)

Vikas TU
13786 Points
							
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
2 months ago
Vikas Amritiya
askIITians Faculty
158 Points
							

 

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=ut21gt2
 
=10×22110×22=10
 
 Height of tower, h=10 m
2 months ago
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