Answer:
u x = 16.04 m/s
Explanation:
given,
height of the hill = 22 m
the soccer ball lands at = 35 m
initial horizontal velocity = ?
using equation of motion
[tex]s = ut +\dfrac{1}{2}at^2[/tex]
[tex]22 = \dfrac{1}{2}\times 9.8 \times t^2[/tex]
[tex]t = \sqrt{\dfrac{2\times 22}{9.8}}[/tex]
t = 2.12 s
[tex]s = ut +\dfrac{1}{2}at^2[/tex]
[tex]35 = u\times 2.12 +\dfrac{1}{2}\times 0 \times 2.12^2[/tex]
35 = 2.12 u
u x = 16.04 m/s
Answer:
The initial horizontal velocity of the soccer ball is 6.25 m/s.
Explanation:
Given data:
Height of hill is, [tex]H=22.0 \;\rm m[/tex].
Distance covered is, [tex]s=35.0 \;\rm m[/tex].
Apply second kinematic equation of motion as,
[tex]H=ut+ \dfrac{1}{2}gt^{2}[/tex]
Here, u is the velocity of ball before kicking, g is gravitational acceleration and t is the time.
[tex]22=0 \times t+ \dfrac{1}{2} \times 9.8 t^{2}\\t= 2.11 \;\rm s[/tex]
Now, again
[tex]d=u' \times t+ \dfrac{1}{2}gt^{2}[/tex]
Here, u' is the initial velocity after kicking. Solving as,
[tex]35=u' \times 2.11+ \dfrac{1}{2} \times 9.8 \times 2.11^{2}\\u' = 6.25 \;\rm m/s[/tex]
Thus, the initial horizontal velocity of the soccer ball is 6.25 m/s.
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https://brainly.com/question/14355103?referrer=searchResults
