Respuesta :
Answer:
Maximum height reached = 35.15 meter.
Explanation:
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g [tex]m/s^2[/tex] and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 [tex]m/s^2[/tex] and time taken = 2u sin θ /g
So range of projectile, [tex]R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}[/tex]
Vertical motion (Maximum height reached, H) :
We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
[tex]0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}[/tex]
In the give problem we have R = 301.5 m, θ = 25° we need to find H.
So [tex]\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g[/tex]
Now we have [tex]H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m[/tex]
So maximum height reached = 35.15 meter.
The maximum height that can reach by ball is 35.15 m.
Given Here,
R = 301.5m
[tex]\rm \bold\theta = 25^\cdot[/tex]
The vertical motion
[tex]\rm \bold{ v=u +at}[/tex]
Where,
V - final velocity = [tex]\rm \bold{ usin \theta}[/tex]
u- initial velocity = 0 m/s
a- acceleration =[tex]\rm \bold{ -g m/s^2}[/tex]
t - time
put the value
[tex]\rm \bold {t =\frac{2u sin\theta}{g} }[/tex]
The range of projectile motion
[tex]\rm \bold { R = \frac{u^2 sin^2 \theta}{g} }[/tex]
[tex]\rm u^2 = 393.59 g\\[/tex]
The height
[tex]\bold {H= \frac{u^2 \times sin^2\theta}{2g} }[/tex]
Putting the values in the equation
H = 35.15 m
Hence, we can conclude that the maximum height that can reach by ball is 35.15 m.
To know more about projectile motion, refer the link:
https://brainly.com/question/11049671?referrer=searchResults
