A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimate the maximum height the ball reaches. (Use 10 m/s2 as the acceleration of gravity.) 101 Incorrect: Your answer is incorrect.

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √ (v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[ (15 m/s) ² + (30 m/s) ²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ) / 2g

H = (33.54 m/s) ² (Sin² 63.43°) / 2 (10 m/s²)

H = 45 m