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A player throws a ball upwards with an initial speed of $29.4 \mathrm{~ms}^{-1}$. (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 the $x=0, t=0$ be the location and time 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 upwards, and downward motion. (d) To what height does the ball rise and after how long does the ball returns to the player's hands. $\left(g=9.8 \mathrm{~ms}^{-2}\right.$ and air resistance is negligible).
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(a) Since the ball is moving under the effect of gravity, the direction of acceleration due to gravity is always vertically downward.
(b) At the highest point, the vertical velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity $=9.8 \mathrm{~ms}^{-2}$ in vertically downward direction.
(c) When the highest point is chosen at the location $x=$ 0 and $t=0$ and vertically downward direction to be the positive direction of $x$-axis and upward direction as negative direction of $x$-axis.
During upward motion, position sign is negative, sign of velocity is negative and sign of acceleration is positive.
During downward motion, position sign is positive, sign of velocity is negative and sign of acceleration is positive.
(d) During upward motion
$$
u=-29.4 \mathrm{~ms}^{-1}, a=9.8 \mathrm{~ms}^{-2}, v=0
$$
As $v^2-u^2=2 a s$
$$
\begin{aligned}
&\Rightarrow 0-(29.4)=2 \times 9.8 \times s \\
&\Rightarrow s=\frac{-(29.4)^2}{2 \times 9.8}=-44.1 \mathrm{~m} \\
&\text { Also } v=u+a t \Rightarrow v-u=a t \\
&\Rightarrow 0-(-29.4)=9.8 t
\end{aligned}
$$
or $t=\frac{29.4}{9.8}=3 \mathrm{~s}$
Total time $=3+3=6 \mathrm{~s}$
[ $\because$ time of ascent $=$ time of descent $]$
(b) At the highest point, the vertical velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity $=9.8 \mathrm{~ms}^{-2}$ in vertically downward direction.
(c) When the highest point is chosen at the location $x=$ 0 and $t=0$ and vertically downward direction to be the positive direction of $x$-axis and upward direction as negative direction of $x$-axis.
During upward motion, position sign is negative, sign of velocity is negative and sign of acceleration is positive.
During downward motion, position sign is positive, sign of velocity is negative and sign of acceleration is positive.
(d) During upward motion
$$
u=-29.4 \mathrm{~ms}^{-1}, a=9.8 \mathrm{~ms}^{-2}, v=0
$$
As $v^2-u^2=2 a s$
$$
\begin{aligned}
&\Rightarrow 0-(29.4)=2 \times 9.8 \times s \\
&\Rightarrow s=\frac{-(29.4)^2}{2 \times 9.8}=-44.1 \mathrm{~m} \\
&\text { Also } v=u+a t \Rightarrow v-u=a t \\
&\Rightarrow 0-(-29.4)=9.8 t
\end{aligned}
$$
or $t=\frac{29.4}{9.8}=3 \mathrm{~s}$
Total time $=3+3=6 \mathrm{~s}$
[ $\because$ time of ascent $=$ time of descent $]$
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