Answer:
a) h = 16.7 m
b) t = 3.4 s
c) d = 31.3 m
d) a = 5.42 m/s²
Explanation:
a) The maximum height can be calculated as follows:
[tex] V_{f_{y}}^{2} = V_{0_{y}}^{2} - 2gh [/tex]
Where:
[tex]V_{f_{y}}[/tex]: is the final speed in "y" direction = 0 (for maximum height)
[tex]V_{0_{y}}[/tex]: is the initial speed in "y" direction
h: is the maximum height
g: is the gravity = 9.81 m/s²
[tex]h = \frac{V_{0_{y}}sin(65)^{2}}{2g} = \frac{(20*sin(65))^{2}}{2*9.81} = 16.7 m[/tex]
b) The time that will take for the girl to catch the ball is:
[tex] h_{f} = h_{0} + V_{0_{y}}*t - \frac{1}{2}gt^{2} [/tex]
By solving the above equation for t we have:
[tex] t = \frac{2V_{0_{y}}}{g} = \frac{2*20*sin(65)}{9.81} = 3.70 s [/tex]
Since the girl runs after the ball 0.30 s later the total time is:
[tex] t = 3.70 - 0.3 = 3.4 s [/tex]
c) The distance in x can be found as follows:
[tex] x = V_{0_{x}}*t = 20*cos(65)*3.7 = 31.3 m [/tex]
d) The acceleration of the girl is:
[tex] x_{f} = x_{0} + V_{0_{x}}*t + \frac{1}{2}at^{2} [/tex]
By solving the above equation for "a" we have:
[tex] a = \frac{2*x_{f}}{t^{2}} = \frac{2*31.3}{(3.4)^{2}} = 5.42 m/s^{2} [/tex]
I hope it helps you!
