Question

4) A basketball is launched at a velocity of 25 m/s in a direction making an angle of 50° upward with the

Answer 1

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 .

$\boxed{\bf \:H_{(max)} \: = \cfrac{u {}^{2} {sin}^{2} \theta }{2g}}$

Substituting all the value in above equation we obtain

$\sf \qquad \: \longrightarrow\:H_{(max)} \: = \cfrac{ {25}^{2} {sin}^{2} {50}^{ \circ} }{2 \times 9.8} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cfrac{625 \times(0.766) {}^{2} }{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cfrac{625 \times 0.586756}{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cfrac{366.7225}{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} = \cancel \cfrac{366.7225}{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_{(max)} =18.71 \: m$

Therefore,

• Maximum height reached by the object is 18.71 meters.