a projectile is fired at an angle of A projectile is fired at an angle of 55.0 d

Question

a projectile is fired at an angle of A projectile is fired at an angle of 55.0 degree above the horizontal with an initial speed of 35.0 m/s. What is the magnitude of the horizontal component of the projectile's displacement at the end of 2 s? How long does it take the projectile to reach the highest point in its trajectory? A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. Determine the initial speed of the projectile. At what angle is the projectile fired (measured with respect to the horizontal)? Approximately how long does it take the projectile to reach the highest point in its trajectory? What is the speed of the projectile when it is at the highest point in its trajectory? What is the acceleration of the projectile when it reaches its maximum height? What is the magnitude of the projectile's velocity just before it strikes the ground? A projectile is fired horizontally with an initial speed of 57 m/s. What are the horizontal and vertical components of its displacement 3.0 s after it is fired? horizontal vertical A) 44 m -29 m B) 170 m zero m C) 170 m -44 m D) 210 m -44 m E) 210 m zero m

Explanation / Answer

horizontal component of velocity = v * cos(theta)

horizontal component of velocity = 35 * cos(55)

speed = distance / time

35 * cos(55) = distance / 2

distance = 2 * 35 * cos(55)

horizontal distance = 40.15 m

time taken by projectile to reach highest point = v * sin(theta) / g

time taken = 35 * sin(55) / 9.8

time taken = 2.92 sec

time taken by projectile to reach highest point = 2.92 sec

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