A football kicker is attempting a field goal from $44 \mathrm {~m}$ out.The ball is kicked and just clears the lower bar with a time of flight of $2.9 \mathrm {~s}$ If the angle of the kick was 45°,what was the initial speed of the ball,assuming no effects of air?
A)21.5 m/s
B)19.7 m/s
C)2.2 m/s
D)39 m/s

Question

Quizplus · Accepted Answer

The initial speed of the ball can be found using the range formula for projectile motion: $R = \frac{v^2 \sin 2\theta}{g}$, where $R$ is the range (44 m), $v$ is the initial speed, $\theta$ is the angle of projection (45°), and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Since $\sin(2 \times 45°) = \sin(90°) = 1$, the formula simplifies to $R = \frac{v^2}{g}$. Rearranging for $v$, we get $v = \sqrt{Rg}$. Substituting $R = 44 \, \text{m}$ and $g = 9.8 \, \text{m/s}^2$, we find $v = \sqrt{44 \times 9.8} \approx 21.5 \, \text{m/s}$.

A Football Kicker Is Attempting a Field Goal From 44 m44 \mathrm {~m}44 m

A Football Kicker Is Attempting a Field Goal From $44 \mathrm {~m}$