Answer :
A. We know that the height (h) = 25 m, and the distance (x) = 135 m.
B. We need to find the time flight first (t), then we can find easily the horizontal velocity ( [tex] v_{x} [/tex]. )
C. Because this is a projectile motions, so horizontal velocity is constant, but the vertical velocity is change because of the gravitational acceleration.
So we can use : [tex]h = \frac{1}{2} g t^{2} [/tex],
Note : Time that will be exist is a time from a land to the takeoff point, that can be called as the time flight.
Then we use, [tex]x = v_{x} t[/tex], to find the horizontal velocity.
D. We know that :
[tex]h = \frac{1}{2} g t^{2} [/tex], so by insert the numbers, and assune that the (g = 10 m/[tex] s^{2} [/tex]), we get that ( t = [tex] \sqrt{5} [/tex]. )
After that by using [tex]x = v_{x} t[/tex], we can get that [tex] v_{x} = 27 \sqrt{5} [/tex]
B. We need to find the time flight first (t), then we can find easily the horizontal velocity ( [tex] v_{x} [/tex]. )
C. Because this is a projectile motions, so horizontal velocity is constant, but the vertical velocity is change because of the gravitational acceleration.
So we can use : [tex]h = \frac{1}{2} g t^{2} [/tex],
Note : Time that will be exist is a time from a land to the takeoff point, that can be called as the time flight.
Then we use, [tex]x = v_{x} t[/tex], to find the horizontal velocity.
D. We know that :
[tex]h = \frac{1}{2} g t^{2} [/tex], so by insert the numbers, and assune that the (g = 10 m/[tex] s^{2} [/tex]), we get that ( t = [tex] \sqrt{5} [/tex]. )
After that by using [tex]x = v_{x} t[/tex], we can get that [tex] v_{x} = 27 \sqrt{5} [/tex]
