i need help with number 6 please, let me know how did i do . Thank You A project
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i need help with number 6 please, let me know how did i do . Thank You
A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30m/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 4 horizontal)? Approximately how long does it takes 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? Whit is the acceleration of the projectile when it reaches its maximum what is the magnitude of the projectile's velocity just before it striker the ground?Explanation / Answer
a)initial speed of the projectile is = sqrt(Vx^2 + Vy^2) = sqrt(900+1600) = 50 m/s b)angle at which the projectile is fired .. let it be (theta) so tan(theta) = 40/30 theta = 53.13 degrees c) Time taken to reach the highest point of the trajectory considering y direction motion v at highest point is 0 and it is going against force of gravity in the downward direction so 0 = 40 - 9.8t t = 4.08 sec d) Speed of the projectile in its highest trajectory = only horizontal component of speed so 30 m/s e)Acceleration of the projectile is 9.8 m/s^2 acting in negative Y Direction at highest point. f) Magnitude of velocity when it strikes down is same as the one with which it was thrown = 50 m/s
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