Answer :
Given:
A player catches the ball 3 seconds after throwing it straight up in air.
Let's find the speed and the maximum height of the ball.
• Speed:
To find the speed, apply the motion formula:
[tex]v=u+at[/tex]Where:
v is the final velocity = 0 m/s
u is the initial velocity
a is the acceleration due to gravity = -9.8 m/s²
t is the time it takes the ball to reach maximum height = 3/2 = 1.5 seconds.
Rewrite the formula for u and solve:
[tex]\begin{gathered} u=v-at \\ \\ u=0-(-9.8)(1.5) \\ \\ u=14.7\text{ m/s} \end{gathered}[/tex]Therefore, the seed at which he threw the ball is 14.7 m/s.
• Maximum height:
To find the maximum height, apply the formula:
[tex]v^2=u^2+2as[/tex]Where:
s is the maxiumum height.
Thus, we have:
[tex]\begin{gathered} 0^2=14.7^2+2(9.8)s \\ \\ -14.7^2=19.6s \\ \\ s=\frac{-14.7^2}{19.6}=\frac{216.09}{19.6} \\ \\ s=11.025\text{ m} \end{gathered}[/tex]Therefore, the ball traveled 11.025 m high.
ANSWER:
• Speed = 14.7 m/s
,• Maximum height = 11.025 m