Respuesta :
Assuming U is the initial velocity, g is the acceleration due to gravity and t is the time of flight, the displacement vector for projectile.
[tex]v = u + i + \frac{1}{2}gt^2 j ...(i)[/tex]
where x = ut is the horizontal displacement
[tex]y = \frac{1}{2}gt^2[/tex] is the vertical dispalcement.
The vertical change in height of the projectile after 2.5s is 30.625m and the horizontal distance the projectile travels is 638.87m
Projectile Motion
To calculate the the vertical change in height after 2.5 seconds.
t = 2.5s
Vertical displacement
[tex]y = \frac{1}{2}gt^2\\ y = \frac{1}{2}9.8 * 2.5^2\\ y = 30.625m[/tex]
b)
Let R be the maximum horizontal distance travelled by projectile.
[tex]R = ut + \frac{1}{2}gt^2[/tex]
u = initial velocity
For horizontal motion, we assume g = 0, t = maximum time of flight.
[tex]R = ut...(iii)[/tex]
Maximum time of flight of horizontal projectile
[tex]T = \sqrt{\frac{2h}{g} }[/tex]
where h represent the height from which the projectile is thrown.
Let's substitute the formula of time of flight into equation (iii)
[tex]R = u\sqrt{\frac{2h}{g} }[/tex]
given that u = 200m/s, h = 50m
[tex]R = u\sqrt{\frac{2h}{g} } \\R = 200\sqrt{\frac{2*50}{10} }\\R = 638.87m[/tex]
The vertical change in height of the projectile after 2.5s is 30.625m and the horizontal distance the projectile travels is 638.87m
Learn more on projectile motion here;
https://brainly.com/question/6672263
#SPJ1
