A) 13.13 m (4.13 m above the top of the ramp)
B) 0.92 s, 30.0 m (14.4 m from the top of the ramp)
Explanation:
A)
The motion of the projectile consists of two independent motions:
- A uniform motion along the horizontal direction (constant horizontal velocity)
- A uniformly accelerated motion along the vertical direction (constant downward acceleration)
To find the maximum height, we just analyze the vertical motion.
The initial vertical velocity of the projectile is given by:
[tex]u_y = u sin \theta = (18)(sin 30^{\circ})=9 m/s[/tex]
Where
u = 18 m/s is the initial velocity
[tex]\theta=30^{\circ}[/tex] is the angle of projection
Since this is a uniformly accelerated motion, we can use the following suvat equation:
[tex]v_y^2-u_y^2=2as[/tex]
where:
[tex]v_y=0[/tex] is the final vertical velocity when the projectile reaches the maximum height
[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration due to gravity (negative because it is downward)
s is the vertical displacement
Re-arranging, we find s:
[tex]s=\frac{v_y^2-u_y^2}{2a}=\frac{0^2-(9)^2}{2(-9.8)}=4.13 m[/tex]
However, this is the vertical displacement above the top of the ramp. We see that the ramp is [tex]d=15.6 m[/tex] long in the horizontal direction, so the height of the ramp is
[tex]h=d tan \theta=(15.6)(tan 30^{\circ})=9m[/tex]
So, the maximum height of the projectile is:
[tex]H=h+s=9+4.13 = 13.13 m[/tex]
B)
To find the time at which the projectile reaches the maximum height, we use another suvat equation:
[tex]v_y=u_y + at[/tex]
where:
[tex]v_y=0[/tex] is the vertical velocity at the maximum height
[tex]a=-9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]u_y = 9 m/s[/tex] is the initial vertical velocity
Solving for t, we find the time:
[tex]t=\frac{v_y-u_y}{a}=\frac{0-9}{-9.8}=0.92 s[/tex]
The horizontal position of the projectile instead is given by the equation for uniform motion:
[tex]x(t)=v_x t[/tex]
where:
[tex]v_x = u cos \theta = (18)(cos 30^{\circ})=15.6 m/s[/tex] is the initial horizontal velocity
Substituting t = 0.92 s, we find:
[tex]x=(15.6)(0.92)=14.4 m[/tex]
So, this is the horizontal distance covered from the top of the ramp at the instant of maximum height; and therefore, the horizontal distance from the beginning of the ramp is
[tex]d=15.6+14.4=30.0 m[/tex]
