# Answer on Mechanics | Relativity Question for tinu

Question #50491

A ball is hit with 30 m/s velocity creating an angle of 20 degree. At the place , where the ball reaches it's maximum height, at that point a fielder who is 5 feet and 8 inches tall is fielding. He jumps 2 m high to catch the ballm the distance of the boundary is 70 m from the hitting place, and another fielder is fielding inside 11 m from boundary.

a) will the 1st fielder be able to catch the ball??

b)if the 1st fielder fails to take the catch by any chance , will the 2nd fielder be able to take the catch?

c) does the height of the 1st fielder is necessary to solve part a) ??

a) will the 1st fielder be able to catch the ball??

b)if the 1st fielder fails to take the catch by any chance , will the 2nd fielder be able to take the catch?

c) does the height of the 1st fielder is necessary to solve part a) ??

Expert's answer

A ball is hit with 30 m/s velocity creating an angle of 20 degree.

At the place , where the ball reaches it's maximum height, at that point a fielder who is 5 feet and 8 inches tall is fielding.

He jumps 2 m high to catch the ballm the distance of the boundary is 70 m from the hitting place,

and another fielder is fielding inside 11 m from boundary.

a) will the 1st fielder be able to catch the ball??

b)if the 1st fielder fails to take the catch by any chance , will the 2nd fielder be able to take the catch?

c) does the height of the 1st fielder is necessary to solve part a) ??

Solution:

Given:

v0 = 30; %m/s

g = 9.81;%m/s^2

alpha=20*pi./180; % in radians

h_fieder=1.723 % in m = 5 feet 8 enches

y = y0 + v0*sin(alpha)*t - g*t^2/2

x = x0 + v0*cos(alpha)*t

For our case x0=y0=0

Let's calculate the time when ball hit to the fround

At this case y = v0*sin(alpha)*t - g*t^2/2 = 0 -> t_hit = t_max = 2.*v0.*sin(alpha)./g

Let's calculate maximum distance for ball:

x = v0*cos(alpha)*t_hit = v0*cos(alpha)*2.*v0.*sin(alpha)./g =

Let's calculate maximum heigth for ball:

y = y0 + v0*sin(alpha)*t - g*t^2/2 where t=tmax/2

ymax = v0.*sin(alpha).*(v0.*sin(2*alpha)./(2*g)) - (g/2).*( v0.*sin(2*alpha)./(2*g) ).^2

h_fieder =

1.7230

t_max = 2.0919 sec

x_max = 58.9713 m

y_max = 5.3464 m

So

a) will the 1st fielder be able to catch the ball?? Answer is no

b)if the 1st fielder fails to take the catch by any chance , will the 2nd fielder be able to take the catch? Answer is yes

c) does the height of the 1st fielder is necessary to solve part a) ?? Answer is no

At the place , where the ball reaches it's maximum height, at that point a fielder who is 5 feet and 8 inches tall is fielding.

He jumps 2 m high to catch the ballm the distance of the boundary is 70 m from the hitting place,

and another fielder is fielding inside 11 m from boundary.

a) will the 1st fielder be able to catch the ball??

b)if the 1st fielder fails to take the catch by any chance , will the 2nd fielder be able to take the catch?

c) does the height of the 1st fielder is necessary to solve part a) ??

Solution:

Given:

v0 = 30; %m/s

g = 9.81;%m/s^2

alpha=20*pi./180; % in radians

h_fieder=1.723 % in m = 5 feet 8 enches

y = y0 + v0*sin(alpha)*t - g*t^2/2

x = x0 + v0*cos(alpha)*t

For our case x0=y0=0

Let's calculate the time when ball hit to the fround

At this case y = v0*sin(alpha)*t - g*t^2/2 = 0 -> t_hit = t_max = 2.*v0.*sin(alpha)./g

Let's calculate maximum distance for ball:

x = v0*cos(alpha)*t_hit = v0*cos(alpha)*2.*v0.*sin(alpha)./g =

Let's calculate maximum heigth for ball:

y = y0 + v0*sin(alpha)*t - g*t^2/2 where t=tmax/2

ymax = v0.*sin(alpha).*(v0.*sin(2*alpha)./(2*g)) - (g/2).*( v0.*sin(2*alpha)./(2*g) ).^2

h_fieder =

1.7230

t_max = 2.0919 sec

x_max = 58.9713 m

y_max = 5.3464 m

So

a) will the 1st fielder be able to catch the ball?? Answer is no

b)if the 1st fielder fails to take the catch by any chance , will the 2nd fielder be able to take the catch? Answer is yes

c) does the height of the 1st fielder is necessary to solve part a) ?? Answer is no

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