Projectile motion: vertical velocity from height, add horizontal component for speed

Steps:

• Use vertical kinematics: $v_y^2 = u_y^2 - 2 a h = 8.0^2 - 2(1.6)(10) = 64 - 32 = 32$, so $v_y = \sqrt{32} \approx 5.7$ m/s upward.
• Horizontal velocity remains constant: $v_x = 4.0$ m/s.
• Total speed: $\sqrt{v_y^2 + v_x^2} = \sqrt{32 + 16} = \sqrt{48} \approx 6.9$ m/s, which rounds to 7.0 m/s.
• Confirm height: starting point to 10 m above, as implied by diagram.

Why B is correct:

• Matches the Pythagorean combination of constant horizontal velocity and vertical velocity from the kinematic equation $v_y^2 = u_y^2 - 2gh$.

Why the others are wrong:

• A: Underestimates speed, perhaps using only vertical velocity or incorrect height.
• C: Overestimates, possibly confusing with initial speed $\sqrt{80} \approx 8.9$ m/s or Earth's g.
• D: Matches horizontal speed only, ignoring vertical component at that height.

Final answer: B