Question #2027f
1 Answer
(i)
(ii)
(iii)
Explanation:
We're asked to find
-
(i) the maximum height obtained by the mass
-
(ii) the time it takes the mass to travel
3.0 "m" horizontally -
(iii) the time(s) when the height of the mass is
4.0 "m"
(i)
Finding the maximum height:
When the particle is at its maximum height, the instantaneous
ul((v_y)^2 = (v_0sinalpha_0)^2 - 2g(Deltay)
where
-
v_y is they -velocity at heightDeltay (which is0 ) -
v_0 is the initial speed (given as25 "m/s" ) -
alpha_0 is the launch angle (given as30^"o" ) -
g = 9.81 "m/s"^2 -
Deltay is the change in height (what we're trying to find)
Plugging in known values:
0 = [(25color(white)(l)"m/s")sin(30^"o")]^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)
color(red)(ulbar(|stackrel(" ")(" "Deltay = 7.96color(white)(l)"m"" ")|)
The maximum height is thus
(ii)
The time it takes to travel
To do this, we can use the equation
ul(Deltax = v_0cosalpha_0t
where
-
Deltax is the change in horizontal position (3.0 "m" ) -
v_0 = 25 "m/s" -
alpha_0 = 30^"o" -
t is the time (what we're trying to find)
Plugging in known values:
3.0color(white)(l)"m" = (25color(white)(l)"m/s")cos(30^"o")t
color(red)(ulbar(|stackrel(" ")(" "t = 0.24color(white)(l)"s"" ")|)
It thus takes
(iii)
The time(s) when the height is
We can now use the equation
ul(Deltay = v_0sinalpha_0t - 1/2g t^2
where
-
Deltay is the change in vertical position (4.0 "m" ) -
v_0 = 25 "m/s" -
alpha_0 = 30^"o" -
t is the time (what we're trying to find) -
g = 9.81 "m/s"^2
Plugging in known values:
4.0color(white)(l)"m" = (25color(white)(l)"m/s")sin(30^"o")t - 1/2(9.81color(white)(l)"m/s"^2)t^2
Using the quadratic formula yields
color(red)(ulbar(|stackrel(" ")(" "t = 0.375color(white)(l)"s"color(white)(l)"and"color(white)(l)t = 2.17color(white)(l)"s"" ")|)
Therefore, the mass has a height of