A ball is thrown from the top of one building toward a tall building 50m away. The initial velocity of the ball is 20m/s with 40.0o above horizontal. When will the ball strike the opposite wall?
Question
A ball is thrown from the top of one building toward a tall building 50m away. The initial velocity of the ball is 20m/s with 40.0o above horizontal. When will the ball strike the opposite wall?
Solution
To solve this problem, we need to break it down into horizontal and vertical components.
Step 1: Find the horizontal and vertical components of the initial velocity.
The horizontal component (Vx) is given by V*cos(θ) = 20m/s * cos(40) = 15.32 m/s.
The vertical component (Vy) is given by V*sin(θ) = 20m/s * sin(40) = 12.85 m/s.
Step 2: Find the time it takes for the ball to hit the opposite wall horizontally.
The horizontal distance (d) is given by the formula d = Vx * t. We can rearrange this to find time: t = d / Vx = 50m / 15.32 m/s = 3.26 s.
So, the ball will hit the opposite wall after approximately 3.26 seconds.
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