Questions: This is passing a basketball. u=10 m / s. θ =41°. Then H= cm.

Solution Steps

• Initial velocity, \( u = 10 \, \text{m/s} \)
• Angle of projection, \( \theta = 41^\circ \)
• Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

• Horizontal component, \( u_x = u \cos \theta \)
• Vertical component, \( u_y = u \sin \theta \)

Using the given values: \[ u_x = 10 \cos 41^\circ \] \[ u_y = 10 \sin 41^\circ \]

\[ u_y = 10 \sin 41^\circ \] \[ u_y \approx 10 \times 0.6561 \] \[ u_y \approx 6.561 \, \text{m/s} \]

The maximum height \( H \) can be found using the formula: \[ H = \frac{u_y^2}{2g} \]

Substitute the values: \[ H = \frac{(6.561)^2}{2 \times 9.8} \] \[ H = \frac{43.06}{19.6} \] \[ H \approx 2.197 \, \text{m} \]

\[ H = 2.197 \, \text{m} \times 100 \, \text{cm/m} \] \[ H \approx 219.7 \, \text{cm} \]

Final Answer