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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 $3\, s$ t

Vedclass · Accepted Answer

$29.4\, m/s$,$44.1\, m$,$39.2\, m$

Vedclass · Answer

(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,$v = 0\, m/s$.Acceleration due to gravity,$g = -9.8\, m/s^2$.Using the equation of motion,$v = u + at$,we have:$0 = u + (-9.8 	imes 3)$$u = 9.8 	imes 3 = 29.4\, m/s$.Hence,the ball was thrown upwards with a velocity of $29.4\, m/s$.$(b)$ Let the maximum height attained by the ball be $h$.Initial velocity during the upward journey,$u = 29.4\, m/s$.Final velocity,$v = 0\, m/s$.Acceleration due to gravity,$g = -9.8\, m/s^2$.Using the equation of motion,$s = ut + \frac{1}{2}at^2$:$h = 29.4 	imes 3 - \frac{1}{2} 	imes 9.8 	imes 3^2 = 88.2 - 44.1 = 44.1\, m$.Hence,the maximum height is $44.1\, m$.$(c)$ The ball attains the maximum height after $3\, s$. After attaining this height,it starts falling downwards.In this case,initial velocity,$u = 0\, m/s$.The position of the ball after $4\, s$ of the throw is given by the distance traveled by it during its downward journey in $4\, s - 3\, s = 1\, s$.Using the equation of motion,$s = ut + \frac{1}{2}at^2$:$s = 0 	imes 1 + \frac{1}{2} 	imes 9.8 	imes 1^2 = 4.9\, m$.Now,the total height is $44.1\, m$.This means the ball is $39.2\, m$ $(44.1\, m - 4.9\, m)$ above the ground after $4\, s$.

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

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