Refer to the diagram shown below.
The time between take off and landing is
[tex]t= \frac{dcos(30^{o}) \, m}{25 \, m/s} =0.0346d \, s[/tex]
The vertical drop is
[tex]h=dsin(30^{o})= \frac{1}{2}gt^{2} \\ 0.5d = 0.5(9.8)(0.0346d)^{2} \\0.5=0.00588d\\d=85.034 \, m[/tex]
Part (a)
Therefore,
t = 0.034685.034 = 2.946 s
Answer: 2.95 s (nearest hundredth)
Part (b)
The length of the jump is d = 85.034 m
Answer: 85.0 m (nearest tenth)
Part (c)
The vertical distance is
h = d sin(30) = 42.52 m
Answer: 42.5 m (nearest tenth)
