Answer:
a. 0.683secs and 2.99secs
b.(x,y)=(23.99,11.3)m/s, (23.99,-11.3)m/s\
c. V=30.0m/s at angle ∝=-36.9 Degree
Explanation:
Data given
Velocity, V=30m/s
Angle,∝=36.9 Degree
The motion describe by the baseball is a projectile motion, the velocity at the x-axis and y-axis are given as
[tex]V_{x}=Vcos\alpha\\ V_{x}=30cos36.9\\ V_{x}=23.99m/s\\V_{y}=Vsin\alpha \\V_{y}=30sin36.9\\ V_{y}=18.01m/s[/tex]
a.To calculate the time at which the baseball was at a height of 10m, we use the equation describing the vertical distance traveled
[tex]y=V_{y}t-\frac{1}{2}gt^{2}\\ y=10m\\10=18.01t-\frac{1}{2}*9.81t^{2}\\10=18.01t-4.9t^{2}\\-4.9t^{2}+18.01t-10[/tex]
solving the quadratic equation using the formula method
[tex]t=\frac{-b±\sqrt{b^{2}-4ac }}{2a} \\a=-4.9, b=18.01,c=-10\\t=\frac{18.01±\sqrt{18.01^{2}-4*(-4.9)(-10) }}{2*(-4.9)} \\t=0.683s, t=2.99s[/tex]
Hence the two times required 0.683secs and 2.99secs
b. Note that no acceleration in the hotizontal component, so the velocity remain the same. at a time t=0.683secs,
[tex]V_{x}=23.99m/s, \\V_{y}=V_{yi}-gt\\V_{y}=18.01-9.8*0.683\\V_{y}=11.3m/s\\[/tex]
at a time t=2.99secs,
[tex]V_{x}=23.99m/s, \\V_{y}=V_{yi}-gt\\V_{y}=18.01-9.8*2.99\\V_{y}=-11.3m/s\\[/tex]
c.The landing velocity is the same as the initial projected velocity but in opposite direction i.e V=30.0m/s at angle ∝=-36.9 Degree
