a ball is thrown downward at a speed of 20 m/s from a height of 500 m. consideri

Question

a ball is thrown downward at a speed of 20 m/s from a height of 500 m. considering g= 10.0 m/s and neglecting the air resistance, answer:
a)is the motion uniform or accelerated?

b)clearly draw the coordinate system (axis) you intend to use; specify the initial velocity (v0) and initial coordinate (y0) in the coordinate system you chose.

c)what is the distance traveled by the ball in the first second of its motion?

d)what is the ball's velocity after one second?

e)How long does it take for the ball to hit the ground?

f)what is the ball's velocity right before it hits the ground?

Explanation / Answer

(a) The motion is accelerates with acceleration a = g = 10.0 m/s (b) Initial velocity is vo = 20 m/s       Initial height is yo = 500 m (c) Distance traveled by the ball in first second is               y = vo*t + (1/2)at2                  = (20 m/s)*(1.0s) + (1/2)(10.0 m/s2)(1.0s)2                  = 25 m (d) The velcoity of the ball after 1.0s is           V = vo + gt               = 20 m/s + 10 m/s2*1.0s               = 30 m/s (e) From the equation of motion             y = vo*t + (1/2)at2             500m = (20 m/s)*t + (1/2)(10 m/s2)t2                         5t2 + 20t - 500 = 0 The solution for the above equation is        t = [- 20 ± (202 - 4*5*(-500)) ] / 2(5) Then we get        t = 8.2s or t = - 12.9s The correct root for the equation is t = 8.2s (f)        Velocity of the ball at the ground is              V2 - vo2 = 2g*yo              V2 = vo2 + 2g*yo Then we get                 v  = 101.98 m/s                                                              = (20 m/s)*(1.0s) + (1/2)(10.0 m/s2)(1.0s)2                  = 25 m (d) The velcoity of the ball after 1.0s is           V = vo + gt               = 20 m/s + 10 m/s2*1.0s               = 30 m/s (e) From the equation of motion             y = vo*t + (1/2)at2             500m = (20 m/s)*t + (1/2)(10 m/s2)t2                         5t2 + 20t - 500 = 0 The solution for the above equation is        t = [- 20 ± (202 - 4*5*(-500)) ] / 2(5) Then we get        t = 8.2s or t = - 12.9s The correct root for the equation is t = 8.2s (f)        Velocity of the ball at the ground is              V2 - vo2 = 2g*yo              V2 = vo2 + 2g*yo Then we get                 v  = 101.98 m/s                                        500m = (20 m/s)*t + (1/2)(10 m/s2)t2                         5t2 + 20t - 500 = 0 The solution for the above equation is        t = [- 20 ± (202 - 4*5*(-500)) ] / 2(5) Then we get        t = 8.2s or t = - 12.9s The correct root for the equation is t = 8.2s (f)        Velocity of the ball at the ground is              V2 - vo2 = 2g*yo              V2 = vo2 + 2g*yo Then we get                 v  = 101.98 m/s                                                  

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