How high would a ball go if it is thrown vertically upward and returns to launching position after 8 seconds?*1 pointA. 7.84 mB. 31.36 mC..78.4 mD. 313.6 m

All are come togetherball is projected from the ground level in vertically upward direction with an initial velocity u. The ball goes up to a maximum height ( EMG ) and then starts moving downward. Finally the ball falls back on the ground. It is given that motion of the ball under the action of gravity alone( IFYS ) and effect of air resistance, if any, is negligible. What is the value of maximum height up to which the ball goes ? (ONT ) What time does the ball take to reach the highest point during its motion ?​​

Every word problem, there should be a conclusion. therefore add a conclusion to the following answer. "(a) The velocity of the ball at its highest point is 0 m/s. This is because at the highest point, the ball stops moving upwards and is about to start falling downwards. (b) The velocity of the ball 1 second before it reaches its highest point depends on the initial speed and the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s². So, if we denote the initial speed as v0, then the velocity 1 second before the highest point is v0 - 9.8 m/s. (c) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is (v0 - 9.8 m/s) - v0 = -9.8 m/s. (d) The velocity of the ball 1 second after it reaches its highest point is -9.8 m/s. This is because the ball has started to fall and is accelerating downwards due to gravity. (e) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is -9.8 m/s - 0 = -9.8 m/s. (f) The change in velocity during the 2-second interval is the final velocity after 2 seconds minus the initial velocity, which is (-9.8 m/s) - (v0 - 9.8 m/s) = -19.6 m/s. (g) The acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity is -9.8 m/s². This is the acceleration due to gravity, and it is always acting on the ball, regardless of its velocity."

A ball thrown up vertically returns to the thrower after 6 s. (a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c) its position after 4 s.

A ball is thrown vertically upward with a velocity of 49 m/s. calculate:(a) The maximum height to which it rises.(b) The total time it takes to return to the surface of the earth. (Take g=9.8m/s2)

A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function =ht−80t16t2. After how long will it reach its maximum height?Do not round your answer.