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A player throws a ball upwards with an initial speed of 29.4ms ?1 . (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the players hands ? (Take g=9.8m/s 2 and neglect air resistance).

Answers (1)


a. Vertically downward
b. v = 0 and a = -9.8 m s-2

c. During upward motion, position is positive, velocity is negative but acceleration is positive. During downward motion, position, velocity and acceleration are positive


v_0=29.4 m / s , v =0 m / s$ and $a =-9 / 8 m / s ^2$ \ Distance can be calculated as follows: \ $v^2=v_0^2+2 a X$ \ Or, $0=(29.4)^2+2 	imes(-9.8) 	imes X$ \ Or, $19.6 X =864.36$ \ Or, $X=frac864.3619.6=44.1 m$ \ Now, time can be calculated as follows: \ $v=u+a t$ \ Or, $0=29.4-9.8 t$ \ Or, $t=frac29.49.8=3 s$ \ Total time (ascent $+$ descent $)=3+3=6 s

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Deependra Verma

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