Answer:
a) v_{p} = 2.83 m / s
, b) 50.5º north east
Explanation:
This is a vector problem.

The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground
To make the sum we decompose the speed of the ball in its components
The angle of 30 east of the south, measured from the positive side of the x axis is
θ = 30 + 270 = 300
=
cos 300
= v_{b} sin. 300
v_{bx} = 3.60 cos 300 = 1.8 m / s
v_{by} = 3.60 sin 300 = -3,118 m / s
Let's add speeds on each axis
X axis
vₓ = v_{bx}
vₓ = 1.8 m / s
Y Axis
= v1 - vpy
v_{y} = 5.30 - 3.118
v_{y} = 2.182 m / s
The magnitude of the velocity can be found using the Pythagorean theorem
= √ (vₓ² + v_{y}²)
v_{p} = √ (1.8² + 2.182²)
v_{p} = 2,829 m / s
v_{p} = 2.83 m / s
b) for direction use trigonometry
tan θ =
/ vₓ
θ = tan ⁺¹ v_{y} / vₓ
θ = tan⁻¹ 2.182 / 1.8
Tea = 50.48º
This address is 50.5º north east
Total displacement along the length of mountain is given as
L = 235 m
angle of mountain with horizontal = 35 degree
now we will have horizontal displacement as
x = L cos35
x = 235 cos35 = 192.5 m
similarly for vertical displacement we can say
y = L sin35
y = 235 sin35 = 134.8 m
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Answer:
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Explanation:
I am trying to solve this Q
but not sure if this a correct answer
or not.
Hope this helps.