Asked by Anne
A projectile hits a target placed on a level field. It has arange of 37.0 m. The nagle of launch is 33.4 degrees above the horizontal. What was the initial speed?
Answers
Answered by
Damon
The horizontal component of velocity was constant so
(V cos 33.4)t = 37
V t = 44.3
V = 44.3/t
in the up direction
y = 0 + V sin 33.4 t - 4.9 t^2
0 = 0 + 24.38 =4.9 t^2
4.9 t^2 = 24.38
t^2 = 4.98
t = 2.23seconds in air
V = 44.3/2.23
V = 19.9 m/s
(V cos 33.4)t = 37
V t = 44.3
V = 44.3/t
in the up direction
y = 0 + V sin 33.4 t - 4.9 t^2
0 = 0 + 24.38 =4.9 t^2
4.9 t^2 = 24.38
t^2 = 4.98
t = 2.23seconds in air
V = 44.3/2.23
V = 19.9 m/s
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