2020-03-01T06:55:04-05:00
Throwing the ball at a horizontal angle α = 45o and throwing it at Vo = 10m / s
Elastic collision of the wall at L = 3m. The point at which the ball hits the wall
position, position of drop point on the ground, ball at the moment of collision with the wall
Find the velocity v, respectively.
1
2020-03-03T10:04:33-0500
1)
t = L v 0 cos 45 = 3 10 cos 45 = 0.424 s t=\frac{L}{v_0\cos{45}}=\frac{3}{10\cos{45}}=0.424\ s t = v 0 cos 45 L = 10 cos 45 3 = 0.424 s
y = v 0 sin 45 t − 0.5 g t 2 y=v_0\sin{45}t-0.5gt^2 y = v 0 sin 45 t − 0.5 g t 2
y = 10 sin 45 ( 0.424 ) − 0.5 ( 9.8 ) ( 0.424 ) 2 = 2.1 m y=10\sin{45}(0.424)-0.5(9.8)(0.424)^2=2.1\ m y = 10 sin 45 ( 0.424 ) − 0.5 ( 9.8 ) ( 0.424 ) 2 = 2.1 m 2)
v x ′ = − v x = − v 0 cos 45 = − 7.07 m s v_x'=-v_x=-v_0\cos{45}=-7.07\frac{m}{s} v x ′ = − v x = − v 0 cos 45 = − 7.07 s m
v y ′ = − v y = − ( v 0 sin 45 − g t ) = − 2.92 m s v_y'=-v_y=-(v_0\sin{45}-gt)=-2.92\frac{m}{s} v y ′ = − v y = − ( v 0 sin 45 − g t ) = − 2.92 s m
− 2.1 = − 2.92 t ′ − 0.5 ( 9.8 ) t ′ 2 -2.1=-2.92t'-0.5(9.8)t'^2 − 2.1 = − 2.92 t ′ − 0.5 ( 9.8 ) t ′2
t ′ = 0.421 s t'=0.421\ s t ′ = 0.421 s
x = L + v x ′ t ′ = 3 − 7.07 ( 0.421 ) = 0.024 m x=L+v_x't'=3-7.07(0.421)=0.024\ m x = L + v x ′ t ′ = 3 − 7.07 ( 0.421 ) = 0.024 m
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