projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. The projectile hits the incline plane at point M.
a) Find the time it takes for the projectile to hit the incline plane.
b)Find the distance OM.
Answers
Answer:
Explanation:
a) The x and y components of the displacement are given by
x = V0 cos(θ) t y = V0 sin(θ) t - (1/2) g t2
with θ = 22 + 10 = 32° and V0 = 15 m/s
The relationship between the coordinate x and y on the incline is given by
tan(10°) = y / x
Substitute x and y by their expressions above to obtain
tan(10°) = ( V0 sin(θ) t - (1/2) g t2) / V0 cos(θ) t
Simplify to obtain the equation in t
(1/2) g t + V0 cos(θ) tan(10°) - V0 sin(θ) = 0
Solve for t
t =
V0 sin(θ) - V0 cos(θ) tan(10°)
0.5 g
=
15 sin(32°) - 15 cos(32°) tan(10°)
0.5 (9.8)
= 1.16 s
b)
OM = √[ (V0 cos(θ) t)2 + ( V0 sin(θ) t - (1/2) g t2)2 ]
OM (t=1.16)= √[ (15 cos(32) 1.16)2 + ( 15 sin(32) 1.16 - (1/2) 9.8 (1.16)2)2 ] = 15 meters
Answer:
The formulas for the components Vx and Vy of the velocity and components x and y of the displacement are given by
Vx = V0 cos(θ) Vy = V0 sin(θ) - g t
x = V0 cos(θ) t y = V0 sin(θ) t - (1/2) g t2
In the problem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2.
The height of the projectile is given by the component y, and it reaches its maximum value when the component Vy is equal to zero. That is when the projectile changes from moving upward to moving downward.(see figure above) and also the animation of the projectile.
Vy = V0 sin(θ) - g t = 0
solve for t
t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds
Find the maximum height by substituting t by 0.86 seconds in the formula for y
maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 meters
b) The time of flight is the interval of time between when projectile is launched: t1 and when the projectile touches the ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence
V0 sin(θ) t - (1/2) g t2 = 0
Solve for t
t(V0 sin(θ) - (1/2) g t) = 0
two solutions
t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g
Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.
