As the ball rises from the ground to its maximum height, its vertical velocity will decrease from 10 m/s to 0 m/s at the rate of 9.8 m/s each second. As the ball falls from its maximum height to the ground, its vertical velocity will decrease from 0 m/s to -10 m/s rate of 9.8 m/s each second. Since the horizontal velocity is constant, we can use the following equation to determine the magnitude of the ball's velocity when it lands.
Speed = √(Vertical^2 + Horizontal^2) = √(-10^2 + 2^2) = √104 This is approximately 10.2 m/s.
Tan θ = Vertical ÷ Horizontal = -10 ÷ 2 = -5 The angle is approximately 78.7˚ below horizontal.
When the ball is on the ground, h = 0. So 0 = (1/2)at² + v(i)t; solve for t.
The ball will be acted upon by the wind for time t. But we really don't have enough info to determine how much of that wind velocity produces horizontal motion of the ball. If could be the full 2 m/s, or it could be much less.
Answers
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the y-component will be -10 m/s
and we'll assume the x-component is +2 m/s
then
V^2 = 10^2 + 2^2 = 104
V = 10.2 m/s
tan(angle) = Vy / Vx = -10 / 2 = -5
angle = -78.6°
that's 78.6° below horizontal
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As the ball rises from the ground to its maximum height, its vertical velocity will decrease from 10 m/s to 0 m/s at the rate of 9.8 m/s each second. As the ball falls from its maximum height to the ground, its vertical velocity will decrease from 0 m/s to -10 m/s rate of 9.8 m/s each second. Since the horizontal velocity is constant, we can use the following equation to determine the magnitude of the ball's velocity when it lands.
Speed = √(Vertical^2 + Horizontal^2) = √(-10^2 + 2^2) = √104
This is approximately 10.2 m/s.
Tan θ = Vertical ÷ Horizontal = -10 ÷ 2 = -5
The angle is approximately 78.7˚ below horizontal.
h = (1/2)at² + v(i)t
When the ball is on the ground, h = 0. So
0 = (1/2)at² + v(i)t; solve for t.
The ball will be acted upon by the wind for time t. But we really don't have enough info to determine how much of that wind velocity produces horizontal motion of the ball. If could be the full 2 m/s, or it could be much less.