Question

without air resistance, a kicked ball would reach a maximum height of 6.7 meters and land 38 meters away. with air resistance, the ball would travel
6.7 m vertically and more than 38 m horizontally
38 m horizontally and less than 6.7 m vertically
more than 6.7 m vertically and less than 38 m horizontally
less than 38 m horizontally and less than 6.7 m vertically
a football is thrown at an angle of 30.° above the horizontal. the magnitude of the horizontal component of the balls initial velocity is 13.0 meters per second. the magnitude of the vertical component of the balls initial velocity is 7.5 meters per second. neglect friction.
60.
on the axes in your answer booklet, draw a graph representing the relationship between the horizontal displacement of the football and the time the football is in the air. 1
61 - 62
the football is caught at the same height from which it is thrown. calculate the total time the football was in the air. show all work, including the equation and substitution with units. 2
t =??
g = 9.81 m/s²
vi = 7.5 m/s
vf = 0 m/s
vf = vi + at
0 = 7.5 + 9.81t
-7.5 / 9.81 = 9.81t / 9.81
t = 0.77s

Explanation:

Step1: Identify the vertical motion equation

The equation for vertical motion is \(v_f = v_i+at\). At the maximum height, \(v_f = 0\). But since the ball is caught at the same height, the total time \(t_{total}\) is twice the time to reach the maximum - height.
The vertical initial velocity \(v_{i,y}=7.5\ m/s\) and the acceleration \(a=-g=- 9.81\ m/s^{2}\) (negative because it acts against the motion).
First, find the time to reach the maximum - height using \(v_f = v_i+at\). Substituting \(v_f = 0\), \(v_i = 7.5\ m/s\) and \(a=-9.81\ m/s^{2}\):
\(0 = 7.5-9.81t\)

Step2: Solve for the time to reach the maximum - height

Rearrange the equation \(0 = 7.5-9.81t\) to solve for \(t\):
\(9.81t=7.5\)
\(t=\frac{7.5}{9.81}\ s\approx0.7645\ s\)

Step3: Calculate the total time

Since the total time \(t_{total}\) (time of flight) is twice the time to reach the maximum - height (\(t\)), \(t_{total}=2t\)
\(t_{total}=2\times\frac{7.5}{9.81}\ s=\frac{15}{9.81}\ s\approx1.53\ s\)

Answer:

The total time the football was in the air is approximately \(1.53\ s\)