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a projectile is fired with an initial speed of 500m/s and angle of elevation 30 degrees. find the range, max height and speed at impact

User Noup
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2 Answers

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Final answer:

To find the range, maximum height, and speed at impact of a projectile fired with an initial speed of 500m/s and angle of elevation 30 degrees, we can use the equations of projectile motion.

Step-by-step explanation:

To find the range, maximum height, and speed at impact of a projectile fired with an initial speed of 500m/s and angle of elevation 30 degrees, we can use the equations of projectile motion.

Range: The horizontal distance covered by the projectile is the range. In this case, we can use the formula:

R = (V02 * sin(2θ)) / g

where R is the range, V0 is the initial speed, θ is the angle of elevation, and g is the acceleration due to gravity.

Substituting the given values into the formula, we have:

R = (5002 * sin(2 * 30)) / 9.8

R = (250000 * sin(60)) / 9.8

R ≈ 1450.83 meters

Maximum Height: The maximum height reached by the projectile can be found using the formula:

H = (V02 * sin2(θ)) / (2 * g)

Substituting the given values into the formula:

H = (5002 * sin2(30)) / (2 * 9.8)

H ≈ 3060.45 meters

Speed at Impact: The speed at impact is the magnitude of the velocity of the projectile when it hits the ground. Since the projectile is fired at an angle and lands at a lower height than its initial position, the speed at impact will be lower than the initial speed. However, we need more information to calculate the exact speed at impact.

User Ruzard
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Vt = Voy - at
Vt = initial speed x sin 30 - 10t

0 = 500 sin30 - 10t
0 = 250√3 - 10t
t = 25 √3

hope this helps
User Irfan Ayaz
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