Question

a ball is thrown at an angle of 45° to the ground. if the ball lands 81 m away, what was the initial speed of the ball? (round your answer to the nearest whole number. use g ≈ 9.8 m/s2.) v0 = m/s

Answer 1

By using the range equation for projectile motion, we find that the initial speed for a ball thrown at a 45° angle and landing 81 m away is approximately 28 m/s after rounding to the nearest whole number.

Step-by-step explanation:

To calculate the initial speed of a ball thrown at a 45° angle that lands 81 m away, we use the range equation for projectile motion:

• R = ₲(V₀² sin 2θ) / g

Where R is the range, V₀ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. Given that the ball lands 81 m away (R), the angle θ is 45°, and g is 9.8 m/s², we can solve for V₀.

Since sin(90°) = 1 and the angle is 45°, the equation simplifies to:

R = V₀² / g

Therefore:

V₀ = sqrt(R × g)

Plugging in the values:

V₀ = sqrt(81 × 9.8)

V₀ ≈ sqrt(793.8)

V₀ ≈ 28.2 m/s

After rounding to the nearest whole number, the initial speed of the ball is approximately 28 m/s.