Question

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

horizontal. What is the maximum height reached by the object?

Answer 1

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 * 9.8} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_((max)) = \cfrac{625 *(0.766) {}^(2) }{19.6} \\ \\ \\ \sf \qquad \: \longrightarrow\:H_((max)) = \cfrac{625 * 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.