Question

5). a football is kicked into the air at an angle so that it is in the air for a total of 2 s and has a range of 42 m.
a). how long does it take the football to just go up?
b). how long does it take the football to travel to the top of its path?
c). what is the initial vertical velocity of the football?
d). what is the initial horizontal velocity of the football?

Explanation:

Step1: Time - up calculation

The time - of - flight of a projectile $T = 2s$. The time taken to go up is half of the total time - of - flight.
$t_{up}=\frac{T}{2}$

Step2: Substitute the value of $T$

Given $T = 2s$, then $t_{up}=\frac{2}{2}=1s$

Step3: Find initial vertical velocity

We use the kinematic equation $v = v_0+at$. At the top - most point, the final vertical velocity $v = 0$, the acceleration due to gravity $a=-g=- 9.8m/s^{2}$ and $t = 1s$. From $v = v_{0y}+at$, we can solve for $v_{0y}$:
$v_{0y}=v - at$. Substituting $v = 0$, $a=-9.8m/s^{2}$ and $t = 1s$, we get $v_{0y}=0-(-9.8)\times1 = 9.8m/s$

Step4: Find initial horizontal velocity

The range of a projectile $R = v_{0x}T$. Given $R = 42m$ and $T = 2s$. We can solve for $v_{0x}$ from the formula $v_{0x}=\frac{R}{T}$
$v_{0x}=\frac{42}{2}=21m/s$

Answer:

a. $1s$
b. $1s$
c. $9.8m/s$
d. $21m/s$