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Sagot :
Answer:
As Per Provided Information
- Velocity of projection u is 25m/s
- Angle made by ball ∅ is 50°
We have been asked to determine the maximum height reached by the object .
here we will take acceleration due to gravity g is 9.8 m/s².
For calculating the maximum height attained by the object we will use the following formula .
[tex] \boxed{\bf \:H_{(max)} \: = \cfrac{u {}^{2} {sin}^{2} \theta }{2g}}[/tex]
Substituting all the value in above equation we obtain
[tex]\sf \qquad \: \longrightarrow\:H_{(max)} \: = \cfrac{ {25}^{2} {sin}^{2} {50}^{ \circ} }{2 \times 9.8} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cfrac{625 \times(0.766) {}^{2} }{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cfrac{625 \times 0.586756}{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cfrac{366.7225}{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cancel \cfrac{366.7225}{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} =18.71 \: m[/tex]
Therefore,
- Maximum height reached by the object is 18.71 meters.
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