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answer c,d,and e

t=9.80m/s220.9m/s= s (b) Determine the stone's maximum height. Substitute the time found in part (a) into Equation (2). (c) Find the time the stone takes to return to its initial position, and find the velocity of the stone at that time. Set y=0 in Equation (2) and solve for t. Substitute the time into Equation (1) to get the velocity. (d) Find the time required for the stone to reach the ground. In Equation (2), set y=58.5m. Apply the quadratic formula and take the positive root. (e) Find the velocity and position of the stone at t=6.66s. Substitute values into Equations (1) and (2). ymax=(20.9m/s)(ts)(4.90m/s2)(ts)2=m 0=(20.9m/s)t(4.90m/s2)t2=t(20.9m/s4.90m/s2t)s I s The response you submitted has the wrong sign. v=20.9m/s+(9.80m/s2)(t)v=0.026Xm/s Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. 58.5m=(20.9m/s)t(4.90m/s2)t2t=s v=(9.80m/s2)(6.66s)+20.9m/sv=44.368m/sy=(20.9m/s)(6.66s)(4.90m/s2)(6.66s)2y=106.56m The response you submitted has the wrong sign. (a) Find the time when the stone reaches its maximum height. Write the velocity and position kinematic equations. v=at+v0y=yy0=v0t+1/2at2 Substitute a=9.80m/s2,v0=20.9m/s, and y0=0 into the preceding two equations. v=(9.80m/s2)t+20.9m/s(1)y=(20.9m/s)t(4.90m/s2)t2(2) Substitute v=0, the velocity at maximum height, into Equation (1) and solve for time. 0=(9.80m/s2)t+20.0m/st=9.80m/s220.9m/s= s (b) Determine the stone's maximum height. Substitute the time found in part (a) into Equation (2). (c) Find the time the stone takes to return to its initial position, and find the velocity of the stone at that time. Set y=0 in Equation (2) and solve for t. 0=(20.9m/s)t(4.90m/s2)t2=t(20.9m/s4.90m/s2t)t=0.45s The response you submitted has the wrong sign. Substitute the time into Equation (1) to get the velocity. v=20.9m/s+(9.80m/s2)(t)v=0.026Xm/s Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. (d) Find the time required for the stone to reach the ground. In Equation (2), set y=58.5m. 58.5m=(20.9m/s)t(4.90m/s2)t2 Apply the quadratic formula and take the Substitute v=0, the velocity at maximum height, into Equation (1) and solve for time. 0=(9.80m/s2)t+20.0m/s t=9.80m/s220.9m/s= s (b) Determine the stone's maximum height. Substitute the time found in part (a) into Equation (2). (c) Find the time the stone takes to retum to its initial position, and find the velocity of the stone at that time. Set y=0 in Equation (2) and solve for t. Substitute the time into Equation (1) to get the velocity. (d) Find the time required for the stone to reach the ground. In Equation (2), sety=58.5m. Apply the quadratic formula and take the positive root. (e) Find the velocity and position of the stone at t=6.66 s. Substitute values into Equations (1) and (2). 0=(20.9m/s)t(4.90m/s2)t2=t(20.9m/s4.90m/s2t)t=0.45s The response you submitted has the wrong sign. v=20.9m/s+(9.80m/s2)(t)v=0.026m/s Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. v=(9.80m/s2)(6.66s)+20.9m/sv=44.368m/sy=(20.9m/s)(6.66s)(4.90m/s2)(6.66s)2y=106.56xm (e) Determine the velocity and position of the stone at t=6.66s. Strategy The diagram in Figure 2.20 establishes a coordinate system with y0=0 at the level of release from the thrower's hand, with y positive upward. Write the velocity and position kinematic equations for the stone and substitute the given information. All the answers come from these two equations by using simple algebra or just substituting the time. For part (a), for example, the stone comes to rest for an instant at maximum height, so set v=0 at this point and solve for time. Then substitute the time into the displacement equation, obtaining the maximum height