Answer:
try 70
Step-by-step explanation:
Answer:
160D^2
Step-by-step explanation:
<h2>
Answer:</h2>
![\boxed{\overline{MN}=37.96}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Coverline%7BMN%7D%3D37.96%7D)
<h2>
Step-by-step explanation:</h2>
For a better understanding of this problem, see the figure below. Our goal is to find
. Since:
![\angle MRS=\angle MQP=90^{\circ} \\ \\ \overline{MQ}=\overline{MR}=30mm](https://tex.z-dn.net/?f=%5Cangle%20MRS%3D%5Cangle%20MQP%3D90%5E%7B%5Ccirc%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BMQ%7D%3D%5Coverline%7BMR%7D%3D30mm)
and
is a common side both for ΔMRN and ΔMQN, then by SAS postulate, these two triangles are congruent and:
![\overline{RN}=\overline{QN}](https://tex.z-dn.net/?f=%5Coverline%7BRN%7D%3D%5Coverline%7BQN%7D)
By Pythagorean theorem, for triangle NQP:
![\overline{QN}=\sqrt{\overline{NP}^2+\overline{QP}^2} \\ \\ \overline{QN}=\sqrt{10^2+21^2} \\ \\ \overline{QN}=\sqrt{541}](https://tex.z-dn.net/?f=%5Coverline%7BQN%7D%3D%5Csqrt%7B%5Coverline%7BNP%7D%5E2%2B%5Coverline%7BQP%7D%5E2%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BQN%7D%3D%5Csqrt%7B10%5E2%2B21%5E2%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BQN%7D%3D%5Csqrt%7B541%7D)
Applying Pythagorean theorem again, but for triangle MQN:
![\overline{MN}=\sqrt{\overline{MQ}^2+\overline{QN}^2} \\ \\ \overline{MN}=\sqrt{30^2+(\sqrt{541})^2} \\ \\ \boxed{\overline{MN}=37.96}](https://tex.z-dn.net/?f=%5Coverline%7BMN%7D%3D%5Csqrt%7B%5Coverline%7BMQ%7D%5E2%2B%5Coverline%7BQN%7D%5E2%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BMN%7D%3D%5Csqrt%7B30%5E2%2B%28%5Csqrt%7B541%7D%29%5E2%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7B%5Coverline%7BMN%7D%3D37.96%7D)
Answer:
6.46
Step-by-step explanation:
h(t) = (1/2)
+ v0t + h0
Origin is at surface of water (h = 0), up is positive.
a = -g = -9.8 m/![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
v0 = 25 m/s
h0 = 50 m
h(t) = -4.9t2 + 25t + 50
Find t when
7 =
+ 25t + 50
- 25t - 43 = 0
t = [1/(2*4.9)] [25 ±√(
+ 4*4.9*43)]
t = (0.102)(25 ± 38.312)
t > 0 ⇒ use only the positive square root
t = (0.102)(25 + 38.312) ≅ 6.46 s
It is 7m above the water at about 6.46 s