A ball is thrown upward with such a velocity v that it returns to the thrower after 3 seconds

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(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Final velocity of the ball at the maximum height, v = 0 Acceleration due to gravity, g=−9.8ms−2 From the first equation of motion, v = u + gt 0=u+(−9.8×3) u=9.8×3=29.4ms−1 Hence, the ball was thrown upwards with a velocity of 29.4ms−1 (b) Let the maximum height attained by the ball be h. Initial velocity during the upward journey, u=29.4ms−1 Final velocity, v = 0 Acceleration due to gravity, g=−9.8ms−2 From the second equation of motion, s=ut+12gt2 h=29.4×3+12×−9.8×(3)2=44.1m (c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards. In this case, Initial velocity, u = 0 Distance travelled by it during its downward journey in remaining 1 s is given by s=0×t+12×9.8×12=4.9m Total height = 44.1 m This means that the ball is 39.2 m (44.1 m - 4.9 m) above the ground after 4 seconds.
(i) Velocity
(ii) Maximum height it reaches
(iii) Position after 4 seconds
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A ball thrown up, vertically returns to the thrower after 6 seconds find:a the velocity with which it was thrown upb the maximum height it reachesc its position after 4 seconds

Solution

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Final velocity of the ball at the maximum height, v = 0 Acceleration due to gravity, g=−9.8ms−2 From the first equation of motion, v = u + gt 0=u+(−9.8×3) u=9.8×3=29.4ms−1 Hence, the ball was thrown upwards with a velocity of 29.4ms−1 (b) Let the maximum height attained by the ball be h. Initial velocity during the upward journey, u=29.4ms−1 Final velocity, v = 0 Acceleration due to gravity, g=−9.8ms−2 From the second equation of motion, s=ut+12gt2 h=29.4×3+12×−9.8×(3)2=44.1m (c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards. In this case, Initial velocity, u = 0 Distance travelled by it during its downward journey in remaining 1 s is given by s=0×t+12×9.8×12=4.9m Total height = 44.1 m This means that the ball is 39.2 m (44.1 m - 4.9 m) above the ground after 4 seconds.

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Answer:

A projectile is any object thrown into space upon which the only acting force is gravity. The primary force acting on a projectile is gravity. Projectile motion is defined as the form of motion that is experienced by an object when it is projected into the air, which is subjected to acceleration due to gravity. The projectile is the object while the path taken by the projectile is known as a trajectory.

Given

Time to reach Maximum height,

t = 6/2 = 3 s.

v = 0 (at the maximum height)

a = – 9.8 m s–²

Since the ball is thrown upwards, the acceleration is negative.