Complete Question
A child throws a ball with an initial speed of 8.00 at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground and experience negligible air resistance. (a) What is the magnitude of the ball's velocity just before it hits the ground? (b) At what angle below the horizontal does the ball approach the ground?
Answer:
[tex]v=9.16m/s[/tex]
[tex]\theta=48.1 \textdegree[/tex]
Explanation:
From the question we are told that
Angle of ball [tex]\angle=40 \textdegree[/tex]
Height of ball h=1m
Generally the equation for vertical component is mathematically given by
[tex]v_y=vsin \theta[/tex]
[tex]v_y=8sin 40 \textdegree[/tex]
[tex]v_y=5.14m/s[/tex]
Generally the equation for horizontal component is mathematically given by
[tex]v_x=vcos \theta[/tex]
[tex]v_x=8cos 40 \textdegree[/tex]
[tex]v_x=6.12m/s[/tex]
Generally the equation for vertical displacement of ball is mathematically given by
[tex]y=1+v-yt-\frac{1}{2}gt^2[/tex]
[tex]0=1+5.14t-\frac{1}{2}*9.87*t^2[/tex]
[tex]t=1.22sec[/tex]
[tex]v_yt=v_yt-gt[/tex]
[tex]5.14-(9.8*1.22)=-6.816m/s[/tex]
[tex]v_yt=-6.816m/s[/tex]
Generally the velocity of ball before it hits the ground is mathematically given by
Magnitude
[tex]v=\sqrt{6.12^2 +(-6.816)^2}\\[/tex]
[tex]v=9.16m/s[/tex]
Direction of ball
[tex]\theta=tan^{-1}(\frac{-6.816}{6.12} )[/tex]
[tex]\theta=48.1 \textdegree[/tex] beneath the horizontal
