V^2 = u^2 - 2gh; where v is the final speed, u is the initial speed, g = 9.8 m/s = 0.54 ft/min is the accerelation due to gravity, h = height.
At the maximum height, the final speed is 0.
u^2 = 2gh = 2 x 0.54 x 36 = 38.58
u = sqrt(38.58) = 6.21 ft/min
Therefore, the stne should be thrown with 6.21 ft/min
Answer:
ΔEDC CNBD
Step-by-step explanation:
The triangles show AAA, which cannot be used to demonstrate congruency
Would it be <span>SOLVE FOR TIME TRAVELED WITH IN 60 MILES @ 12MILES/HR
DISTANCE=VELOCITY X TIME
T=S/V
T=60/12
T=5HRS
TOTAL TIME OF JOURNEY =5+3=8HRS.....ANSWER</span>
Answer:
ΔPTS≅ΔRTA by AAS axiom of congruency
Step-by-step explanation:
Consider ΔPQA and ΔRQS
∠PQA=∠RQS (Vertically Opposite Angles)
∠QAP=∠QSR (Complementary of two equal angles, ∠RAT and∠PST)
Due to angle sum property of a triangle, we come to the conclusion that
∠APQ=∠SRQ
Consider ΔPTS and ΔRTA
TA=TS (Given)
∠RAT=∠PST(Given)
∠APQ=∠SRQ (Proved above)
Therefore, ΔPTS≅ΔRTA by AAS axiom of congruency.
Would it be 4(1)^2+2(2(1)+2)^2=100
