Answer :
The base of table is "1.08856 m" far away from the ball land.
Given:
Distance travelled by ball,
- S = 1.2 m
Initial velocity,
- u = 0 m/s
Acceleration,
- a = 9.8 m/s²
Constant speed,
- 2.2 m/s
As we know the formula,
→ [tex]S = ut+\frac{1}{2}at^2[/tex]
By substituting the values, we get
→ [tex]1.2=\frac{1}{2}\times 9.8t^2[/tex]
→ [tex]1.2 = 4.9 t^2[/tex]
→ [tex]t = \sqrt{\frac{1.2}{4.9} }[/tex]
→ [tex]= 0.4948 \ seconds[/tex]
Let,
- Horizontal distance will be "x".
- Time = t
→ [tex]x = 2.2\times t[/tex]
[tex]= 2.2\times 0.4948[/tex]
[tex]= 1.08856 \ m[/tex]
Thus the solution above is right.
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