Solving for Angle of Elevation: p = 12.67°

[thereddevils]

Homework Statement

A small ball is projected with a speed of 10 m/s at an angle of elevation of p from the top of a vertical pole whose height is 15 m from the level ground. The ball hits the ground at a point whose horizontal distance from the foot of the pole is 20 m . Find the value of p . gravity=10 m/s^2

Homework Equations

The Attempt at a Solution

Consider horizontal motion , s=Vx t

20=10 (cos p) t

t=2/(cos p) --1

Consider vertical motion,

15=10 (sin p)(2/(cos p))+1/2 (10)(2\(cos p))^2

15 cos^2 p -20 sin p cos p -20 =0

3 cos 2p - 4 sin 2p=1

5 cos (2p+53.13)=1

2p+53.13=78.46

therefore , p=12.67 degrees

Am i correct ?

 

[rl.bhat]

15=10 (sin p)(2/(cos p))-1/2 (10)(2\(cos p))^2

Till the above step you are right except one sign. Next

15 = 20tan(p) - 20sec^2(p)

Put sec^2(p) = 1 + tan^2(p) and solve for tan(p)

 

[thereddevils]

15=10 (sin p)(2/(cos p))-1/2 (10)(2\(cos p))^2

Till the above step you are right except one sign. Next

15 = 20tan(p) - 20sec^2(p)

Put sec^2(p) = 1 + tan^2(p) and solve for tan(p)

Thanks but isn't the ball being projected downwards which is in the path of the gravity pull ?

 

[rl.bhat]

Thanks but isn't the ball being projected downwards which is in the path of the gravity pull ?

Projected at an angle of elevation means up against the gravitational pull.

 

[americanforest]

Be careful with your signs in your equation for the vertical position. We can use, since the acceleration due to gravity is approximately constant:

$$\Delta y \equiv y_{f}-y_{0}=v_{y}t+\frac{1}{2}at^{2}$$

If we define our coordinate system so that up is positive and down, negative then we are given

$$y_{0}=+15, v_{y}=+10sin(p), a=-g=-9.8$$

Since it starts out above the ground and is shot with an initial velocity up and gravity pulls it down.

 

[thereddevils]

Projected at an angle of elevation means up against the gravitational pull.

got it , thanks