Let v0=25 ms−1, α0=35°,h=20 m, g=9.8 ms−2
a) Let's write an equation for vertical direction:
y=v0sin(α0)t−2gt2y=−h⇒2gt2−v0sin(α0)t−h=0t=gv0sinα0±(v0sinα0)2+2gh
Since time should be ≥0 , we select '+' sign:
t=gv0sinα0+(v0sinα0)2+2gh
t=9.825⋅sin35°+(25⋅sin35°)2+2⋅9.8⋅20 s≈4.0 s
b) We can find velocity magnitude at impact from the law of conservation of energy:
2mv02=2mv2−mgh⇒v=v02+2ghv=252+2⋅9.8⋅20 ms−1≈32 ms−1
The angle between velocity and horizontal we can find using fact that horizontal component of speed remains constant:
vx=v0cosα0=constcosα=vvx=v02+2ghv0cosα0⇒α=±arccosv02+2ghv0cosα0
Since velocity is directed downward, we select '-' sign:
α=−arccos3225⋅cos35°≈−50°
Answers:
a) t≈4.0 s
b) v≈32 ms−1, α≈−50°
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