Question

cricket ball is hit with a velocity 25 m per second 60 m above the horizontal how far above the ground passes over a filter 50m from the bet

Answer 1

Correction : filter = Fielder

bet =bat

Angle from Horizontal: $60\degree$

We are assuming that ball will move in the line-joining the batsman and the fielder mentioned.

Steps:

1) First,we will find time taken to cover 50m along horizontal direction .

So,Components of velocity :

$u_x= 25 cos(60\degree) \\ u_y=25sin(60\degree)$

We know,

$50 = u_x*t \\ => 50 =\frac{25}{2}*t \\ =>t=4s$

2) Now,we will find Displacement in y-direction .

$S_y=u_y t-\frac{1}{2}gt^{2} \\ => \frac{25\sqrt{3}}{2}*4- \frac{1}{2}*10*4^{2} \\ => 50\sqrt{3}- 80 = 6.602\:m (approx)$

So,Ball will at height 6.602m from the fielder mentioned at 50m from the Batsman .

Answer 2

Explanation:

We are assuming that ball will move in the line-joining the batsman and the fielder mentioned.

Steps:

1) First,we will find time taken to cover 50m along horizontal direction .

So,Components of velocity :

\begin{gathered}u_x= 25 cos(60\degree) \\ u_y=25sin(60\degree)\end{gathered}

u

x

=25cos(60°)

u

y

=25sin(60°)

We know,

\begin{gathered}50 = u_x*t \\ = > 50 =\frac{25}{2}*t \\ = > t=4s\end{gathered}

50=u

x

∗t

=>50=

2

25

∗t

=>t=4s

2) Now,we will find Displacement in y-direction .

\begin{gathered}S_y=u_y t-\frac{1}{2}gt^{2} \\ = > \frac{25\sqrt{3}}{2}*4- \frac{1}{2}*10*4^{2} \\ = > 50\sqrt{3}- 80 = 6.602\:m (approx)\end{gathered}

S

y

=u

y

t−

2

1

gt

2

=>

2

25

3

∗4−

2

1

∗10∗4

2

=>50

3

−80=6.602m(approx)

So,Ball will at height 6.602m from the fielder mentioned at 50m from the Batsman .