Question

A ball is thrown at an angle of 45° to the ground. if the ball lands 86 m away, what was the initial speed of the ball? (round your answer to the nearest whole number. use g ≈ 9.8 m/s².)

Answer 1

The initial speed of the ball is approximately 39 m/s.

Step-by-step explanation:

To find the initial speed of the ball, we can use the horizontal and vertical components of velocity. The horizontal distance the ball travels is 86 m, which is the range. The formula for range is:

Range = initial velocity * time of flight

Since the ball is thrown at an angle of 45°, the time of flight is equal to the time it takes for the ball to reach its maximum height and then fall back down. At the maximum height, the vertical component of velocity is zero, so we can determine the time of flight using:

Time of flight = 2 * (vertical component of velocity) / g

Using the given information, we can calculate the initial speed:

Initial speed = Range / (Time of flight)

Substituting the values, we get:

Initial speed = 86 m / (2 * (initial speed * sin(45°)) / 9.8 m/s²)

Simplifying the equation, we get:

Initial speed = 86 m * 9.8 m/s² / (2 * sin(45°))

Calculating the value, the initial speed is approximately 39 m/s.