Answer:
a)40.5m
b)139.5m
Step-by-step explanation:
Considering downward as 'positive' y-direction
[tex]x_i=0\\y_i=0\\v_i=8.8m/s\\a_x=0\\a_y=g\\\theta = 23\degree\\t=5s[/tex]
First is to resolve initial velocity into x and y component:
[tex]v_x_i=v_icos\theta = 8.8 \times cos(23)= 8.1m/s\\v_y_i=v_isin\theta = 8.8 \times sin(23)= 3.4m/s\\[/tex]
a) in order to find horizontal position of the ball:
[tex]x_f=x_i+v_x_it+\frac{1}{2} a_xt^2[/tex]
[tex]x_f= 8.1\times5=40.5m[/tex]
b) To find the height that the ball was thrown from(vertical position):
[tex]y_f=y_i+v_y_it+\frac{1}{2} a_yt^2\\y_f=(3.4\times 5)+(\frac{1}{2} \times9.8\times5^2)\\y_f=139.5m[/tex]
