Asked by Felix
A Stone propelled from a catapult with a speed of 50m/s attains a height of 100m. Calculate the following 1 The time flight, 2 The angle of projection, 3 The. range.
Answers
Answered by
R_scott
1 ... time up equals time down equals T ...100 = 1/2 g T^2
... the flight time is ... 2 T
2 ... 50 sin(Θ) = g T
3 ... range = 2 * T * 50 * cos(Θ)
... the flight time is ... 2 T
2 ... 50 sin(Θ) = g T
3 ... range = 2 * T * 50 * cos(Θ)
Answered by
Anonymous
vertical problem first
No need for anything but g = 9.81 m/s^2 and max height is 100 m
We will get the initial vertical speed Vi
v = Vi - 9.81 t
v = 0 at top
Vi = 9.81 t
h = Hi + Vi t - 4.9 t^2
100 = 0 + 9.81 t^2 - 4.9 t^2
4.9 t^2 = 100
t = 4.52 seconds upward (total of 9.04 in air)
Vi = 9.81*4.52 = 44.3 m/s initial speed upward
so
sin theta = 44.3 / 50
theta = 62.4deg above horizontal
==========================
Horizontal problem
goes at 50 cos 62.4 for 9.04 seconds
u = 23.2 m/s and x = 209 meters
No need for anything but g = 9.81 m/s^2 and max height is 100 m
We will get the initial vertical speed Vi
v = Vi - 9.81 t
v = 0 at top
Vi = 9.81 t
h = Hi + Vi t - 4.9 t^2
100 = 0 + 9.81 t^2 - 4.9 t^2
4.9 t^2 = 100
t = 4.52 seconds upward (total of 9.04 in air)
Vi = 9.81*4.52 = 44.3 m/s initial speed upward
so
sin theta = 44.3 / 50
theta = 62.4deg above horizontal
==========================
Horizontal problem
goes at 50 cos 62.4 for 9.04 seconds
u = 23.2 m/s and x = 209 meters
Answered by
Anonymous
To felix,Plsss number one explain the formulae,and r-scott plss explain in details
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