Question

Projectile Motion: A hobbyist launches a projectile from ground level on a horizontal plain. It reaches a maximum height of 72.3 m and lands 111 m from the launch point, with no appreciable air resistance. What was the angle of launch if g = 9.80 m/s2?

Answer 1

Answer:

The angle of launch from the horizontal direction is 20.99° .

Explanation:

Let u and θ be the initial speed and angle of projection from the horizontal axis of the object respectively.

The equations for projectile motion are :

H = ( u² sin²θ)/ 2g      ......(1)

Here H is maximum height of the projectile motion and g is acceleration due to gravity.

R = ( u² sin2θ)/g         .......(2)

Here R is the maximum horizontal displacement of the object.

Rearrange equation (1) in terms of u².

u² = (2gH)/sin²θ

Substitute this equation in equation (2).

R = (2gH sin2θ) / (sin²θ x g)

R = (2H sin2θ)/sin²θ

Using trigonometry property, sin2θ = 2 cosθ sinθ

So, above equation becomes,

R = (2H x 2 cosθ sinθ)/sin²θ

R = (4H cosθ)/sinθ

tanθ = R/4H

θ = tan⁻¹(R/4H)

Substitute 111 m for R and 72.3 m for H in the above equation.

θ = tan⁻¹( 111/ 4 x 72.3 )

θ = tan⁻¹(0.38)

θ = 20.99°

Answer 2

The angle at which the projectile was launched is 69°.

From the question given above, the following data were obtained:

• Maximum height (H) = 72.3 m

• Range (R) = 111 m

• Acceleration due to gravity (g) = 9.8 m/s²

• Angle (θ) =?

H = u²sine²θ / 2g

72.3 = u²sine²θ / (2 × 9.8)

72.3 = u²sine²θ / 19.6

Cross multiply

u²sine²θ = 72.3 × 19.6

u²sine²θ = 1417.08

Divide both side by sine²θ

u² = 1417.08 / sine²θ ............ (1)

R = u²sine2θ / g  

111 = u²sine2θ / 9.8

Cross multiply

u²sine2θ = 111 × 9.8

u²sine2θ = 1087.8

Divide both side by sine2θ

u² = 1087.8 / sine2θ .............. (2)

Equating equation 1 and 2, we have

1417.08 / sine²θ = 1087.8 / sine2θ

Cross multiply

1417.08 × sine2θ = 1087.8 × sine²θ

Recall:

sine²θ = sineθ × sineθ

sine2θ = 2sineθcosθ

1417.08 × sine2θ = 1087.8 × sine²θ

1417.08 × 2sineθcosθ = 1087.8 × sineθ × sineθ

2834.16 × sineθcosθ = 1087.8 × sineθ × sineθ

Divide both side by sineθ

2834.16 × cosθ = 1087.8 × sineθ

Divide both side by cosθ

2834.16 = 1087.8 × tanθ

Divide both side by 1087.8

Tan θ = 2834.16 / 1087.8

Take the inverse of Tan

θ = Tan¯¹ (2834.16 / 1087.8)

θ = 69°

Thus, the angle at which the projectile was launched is 69°.

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