Given data: Height of table is: h = 0.60 m Horizontal velocity is: v=2.4 m/s Consider the figure as shown below: h The vertical component of the ball's initial velocity is zero i.e., v₂ = 0. x The time required for the pool ball to fall on the ground is: 1 h=v₁t+ = gt² 0.60 m = (0)t + - (9.8 m/s²)ț² (4.9 m/s²)ť² = 0.60 m t = 0.349927 s t≈ 0.350 s The horizontal distance between the table's edge and the ball's landing location is: x=vt = (2.4 m/s)(0.350 s) = 0.840 m
Given data: Height of table is: h = 0.60 m Horizontal velocity is: v=2.4 m/s Consider the figure as shown below: h The vertical component of the ball's initial velocity is zero i.e., v₂ = 0. x The time required for the pool ball to fall on the ground is: 1 h=v₁t+ = gt² 0.60 m = (0)t + - (9.8 m/s²)ț² (4.9 m/s²)ť² = 0.60 m t = 0.349927 s t≈ 0.350 s The horizontal distance between the table's edge and the ball's landing location is: x=vt = (2.4 m/s)(0.350 s) = 0.840 m
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Transcribed Image Text:Given data:
Height of table is: h= 0.60 m
Horizontal velocity is: v= 2.4 m/s
Consider the figure as shown below:
h
The vertical component of the ball's initial velocity is zero i.e., v₁, = 0.
X
The time required for the pool ball to fall on the ground is:
h=v₁t+
+ 17/181²2
0.60 m = (0)t+
+1 (9.8 m/s²)+²
(4.9 m/s² )t² = 0.60 m
t = 0.349927 s
t≈ 0.350 s
The horizontal distance between the table's edge and the ball's landing location is:
x = vt
= (2.4 m/s)(0.350 s)
= 0.840 m
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SOLUTION PROVIDED
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