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A projectile is fired with an initial speed of 30 m/s at an angle of elevation of 30°. Find the following: anionalco at which the maximum height is achieved is The maximum height achieved by the projectile is The time when the projectile hits the ground is 앓 . The range of the projectile is m. The speed of the projectile on impact with the ground is 嚣| m/s.

Answer :

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Answer:

Maximum time at which the height is achieved = 3.06 seconds

b)Maximum height achieved by the projectile = 11.48 m

c) Time when the projectile hits the ground is 1.53seconds

d) Speed of the projectile on impact with the ground is 25.98m/s

Step-by-step explanation:

Given Vo = 30m/s

Vyo = Vo sin theta = 30 × sin30°

Vyo= 30 × 0.5 = 15m/s

Vy = Vyo - gt but Vy=0

0 = 30 - 9.8t

t = 30/ 9.8 = 3.06 seconds

b) Maximum height = zu^2Sin^2thets/2g = 30^2 (0.5)^2/19.6

Maximum height = 225/19.7= 11.48m

c) Time for the projectile to hot the ground =T = sqrt(2×11.48)/9.8

T = sqrt2.343 = 1.53 secs

d) Maximum range R = U^2 Sin2theta/g

R= 30^2×0.5/9.8

R = 450/9.8 = 49.92m

e)

Speed of projectile = Ucos30= 30×0.5=25.98m

Answer:

1) Time to reach maximum height, t = 1.531s

2) Maximum height reached, H = 11.48m

3) Total time of flight, T = 3.062s

4) Range of the projectile, R = 79.53m

5) Speed with which the projectile hits the ground = 30 m/s

Step-by-step explanation:

Given,

U = 30m/s, θ = 30°, g = 9.8 m/s²

For projectile motion, the formulas for the parameters of this motion are fairly popularly know as the following

1) Time to reach maximum height, t, is half of the total time of flight T

t = (u sin θ)/g = (30 × sin 30°)/9.8 = 1.531s

2) Maximum height reached, H

H = (u² sin² θ)/2g = (30² sin² 30°)/(2 × 9.8) = 11.48m

3) Time when the projectile hits the ground is the total time of flight, T,

T = 2t = (2u sin θ)/g = 3.062s

4) The range of the projectile is the horizontal distance it covers.

R = (u² sin 2θ)/g = (30² sin 60°)/9.8 = 79.53m

5) Speed of the projectile as it hits the ground, Vf

Vf = √((Vᵧ)² + Vₓ²)

The final velocity component in the y-direction, Vᵧ is calculated from the equations of motion, using the vertical component of the initial velocity for initial velocity.

Uᵧ = u sin θ = 30 sin 30° = 15 m/s

Vᵧ = 15 - (9.8×3.062) = -15.076 m/s

The horizontal component of the final velocity, Vₓ

Vₓ = u cos θ = 30 cos 30 = 25.98 m/s

Vf = √((-15.076)² + (25.98)²) = 30 m/s.

Hope this Helps!!!

An image of the variation in velocity as the projectile moves is attached to this solution.

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