Answer:
4.17 m/s
Explanation:
To solve this problem, let's start by analyzing the vertical motion of the pea.
The initial vertical velocity of the pea is
[tex]u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s[/tex]
Now we can solve the problem by applying the suvat equation:
[tex]v_y^2-u_y^2=2as[/tex]
where
[tex]v_y[/tex] is the vertical velocity when the pea hits the ceiling
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity
s = 1.90 is the distance from the ceiling
Solving for [tex]v_y[/tex],
[tex]v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s[/tex]
Instead, the horizontal velocity remains constant during the whole motion, and it is given by
[tex]v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s[/tex]
Therefore, the speed of the pea when it hits the ceiling is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s[/tex]
