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 )

Question

Vikas TU · Answer

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 · Answer

The ball is thrown at an angle, θ = 3 0 o . Initial velocity of the ball, u = 1 0 m / s Horizontal range of the ball, R = 1 7 . 3 m We know that, R = u c o s θ t , where t is the time of flight ⟹ t = 5 3 ​ 1 7 . 3 ​ = 2 s e c s using equation of motion we get:- − h = u t − 2 1 ​ g t 2 = 1 0 × 2 − 2 1 ​ 1 0 × 2 2 = − 1 0 ⟹ Height of tower, h = 1 0 m

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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)

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)

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