Final answer:
To calculate the initial speed of the ball thrown at 45° and landing 90 m away, we use the range formula, and after rearranging and solving for the initial speed, we find it to be approximately 30 m/s when rounded to the nearest whole number.
Step-by-step explanation:
To determine the initial speed of the ball when a ball is thrown at an angle of 45° to the ground and the ball lands 90 m away, we can use the following equations for projectile motion:
- Range (R) = (v² · sin(2°)) / g
- Time of flight (T) = 2 · v · sin(°) / g
Given that the range R is 90 m, the angle ° is 45° (which makes sin(2°) = 1), and acceleration due to gravity g is approximately 9.8 m/s², we can rearrange the range equation to solve for the initial speed (v):
R = v² / g
90 m = v² / 9.8 m/s²
v² = 90 m · 9.8 m/s²
v² = 882 m²/s²
v = √(882 m²/s²)
v ≈ 29.7 m/s, rounded to the nearest whole number is 30 m/s.
Therefore, the initial speed of the ball was approximately 30 m/s.