Question

a football is kicked from the ground with a velocity of 40 m/s at an angle of 36.9°. part a how long is the football in the air? t = value units part b how far away does it land? x = value units

Explanation:

Step1: Find vertical - initial velocity

The initial velocity $v_0 = 40$ m/s and the angle $\theta=36.9^{\circ}$. The vertical - initial velocity $v_{0y}=v_0\sin\theta$. So $v_{0y}=40\sin(36.9^{\circ})$. Using $\sin(36.9^{\circ})\approx0.6$, we get $v_{0y}=40\times0.6 = 24$ m/s.

Step2: Calculate time - of - flight (Part A)

The equation for vertical displacement $y = v_{0y}t-\frac{1}{2}gt^{2}$. When the football returns to the ground ($y = 0$), $0=v_{0y}t-\frac{1}{2}gt^{2}=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the initial time). The other is $t=\frac{2v_{0y}}{g}$. Taking $g = 9.8$ m/s², $t=\frac{2\times24}{9.8}\approx4.9$ s.

Step3: Find horizontal - initial velocity

The horizontal - initial velocity $v_{0x}=v_0\cos\theta$. Using $\cos(36.9^{\circ})\approx0.8$, $v_{0x}=40\times0.8 = 32$ m/s.

Step4: Calculate horizontal range (Part B)

The horizontal motion is a uniform - motion with $x = v_{0x}t$. Substituting $v_{0x}=32$ m/s and $t = 4.9$ s, we get $x=32\times4.9 = 156.8$ m.

Answer:

Part A:
$t = 4.9$ s
Part B:
$x = 156.8$ m