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  •  A projectile is launched from the top of a tower of height h with an initial speed of 10 m/s at an angle of 30° above the horizontal. The projectile hits a point on the ground a

 A projectile is launched from the top of a tower of height h with an initial speed of 10 m/s at an angle of 30° above the horizontal. The projectile hits a point on the ground at a horizontal distance of 20 m from the tower. What is the height of the tower?

Option: 1

 10 m 

 


Option: 2

15 m


Option: 3

20 m

 


Option: 4

 25 m


Answers (1)

best_answer

We can use the kinematic equations of motion to solve this problem.

We'll start by finding the time it takes for the projectile to hit the ground.

First, we'll resolve the initial velocity into its horizontal and vertical components:

v_x=10\cos(30)=8.66m/s

v_y=10\sin(30)=5m/s     

Using the vertical motion equation, we can find the time it takes for the projectile to hit the ground:

h=v_yt+0.5gt^{2}

0=5t+0.5*(-9.81)*t^{2}

t=1.02s

Next, we can use the horizontal motion equation to find the horizontal distance the projectile travels:

 x=v_xt

x=8.661.02

x=8.83m       

The horizontal distance travelled is less than the given distance of 20 m, which means the projectile hits the ground at a point that is 20 - 8.83 = 11.17 m  away from the tower.

Now, we can use similar triangles to find the height of the tower. The height of the tower is the distance from the base of the tower to the point where the projectile hits the ground as the vertical component of the initial velocity is to the horizontal component of the initial velocity:

\frac{h}{20-8.83}=\frac{5}{8.66}

 h=15m    

Therefore, the height of the tower is 15 m, which is an option (b).

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