Asked by sandy
A long jumper leaves the ground at an angle of 24.4◦to the horizontal an data speed of 9.52m/s.How far does he jump?The acceleration due to gravity is 9.8m. What maximum height does he reach?
Answers
Answered by
Damon
vertical problem first
Vi = 9.52 sin 24.4
v = Vi - 9.8 t
at top v = 0
so
9.8 t = 9.52 sin 24.4 = 3.93
t = .401 seconds to top
height
h = Vi t - 4.9 t^2
h = 3.93 (.401) - 4.9(.401)^2
h = .789 m
now the horizontal problem
time = 2 t = .802 seconds because he spends as much time falling as rising (same g)
u = 9.52 cos 24.4 forever
u = 8.67
x = u*time = 8.67 * .802 = 6.95 meters
Vi = 9.52 sin 24.4
v = Vi - 9.8 t
at top v = 0
so
9.8 t = 9.52 sin 24.4 = 3.93
t = .401 seconds to top
height
h = Vi t - 4.9 t^2
h = 3.93 (.401) - 4.9(.401)^2
h = .789 m
now the horizontal problem
time = 2 t = .802 seconds because he spends as much time falling as rising (same g)
u = 9.52 cos 24.4 forever
u = 8.67
x = u*time = 8.67 * .802 = 6.95 meters
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