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:
Step-by-step 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:
Step-by-step 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