A projectile is fired with an initial speed of 67.2 m/a at an angle of 36.4 degr

Question

A projectile is fired with an initial speed of 67.2 m/a at an angle of 36.4 degree above the horizontal on a long flat firing range. Part A) Setermine the maximum height reached by the projectile. Part B) determine the total time in the air Part C) determine the total horizontal distance covered ( that is, the range)
Part D) Determine the speed of the projectile 1.00 s after firing .
Part E) Determine the direction of the projectile 1.00 s after firing A projectile is fired with an initial speed of 67.2 m/a at an angle of 36.4 degree above the horizontal on a long flat firing range. Part A) Setermine the maximum height reached by the projectile. Part B) determine the total time in the air Part C) determine the total horizontal distance covered ( that is, the range)
Part D) Determine the speed of the projectile 1.00 s after firing .
Part E) Determine the direction of the projectile 1.00 s after firing Part A) Setermine the maximum height reached by the projectile. Part B) determine the total time in the air Part C) determine the total horizontal distance covered ( that is, the range)
Part D) Determine the speed of the projectile 1.00 s after firing .
Part E) Determine the direction of the projectile 1.00 s after firing

Explanation / Answer

Here,

initial speed , u = 67.2 m/s

theta = 36.4 degree

part A)

maximum height = u^2 * sin^2(theta)/(2 * g)

maximum height = 67.2^2 * sin^2(36.4 degree)/(2 * 9.8)

maximum height = 81.13 m

the maximum height is 81.13 m

part B)

total time in air = 2 * v * sin(theta)/g

total time in air = 2 * 67.2 * sin(36.4 degree)/9.8

total time in air = 8.14 s

the total time in air is 8.14 s

part C)

total horizontal distance covered = t *u * cos(theta)

total horizontal distance covered = 8.14 * 67.2 * cos(36.4 degree)

total horizontal distance covered = 440 m

the total horizontal distance covered is 440 m

part D)

after 1 s

v = 67.2 * cos(36.4 degree) i + (67.2 * sin(36.4 degree) - 9.8) j

v = 54.1 + 30.1 j

v = sqrt(54.1^2 + 30.1^2) = 61.9 m/s

the speed of projectile after 1 s is 61.9 m/s

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