Answer:
A. 18.21 m/s or 65.6 km/h
B. 135.5 km/h
Step-by-step explanation:
A. The vertical speed of the grenade is ...
vv = v·sin(45°) = v/√2
The time in air is given by the equation for ballistic motion:
h = -4.9t² +vv·t = 0
t(vv -4.9t) = 0 . . . . . factor out t
v/√2 = 4.9t
v = t·4.9√2 . . . . . . solve for v
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The horizontal distance the grenade travels in t seconds is ...
d = vh·t = v·cos(45°)·t = vt/√2 = (t·4.9√2)(t/√2) = 4.9t² . . . . meters
1 m/s = 3.6 km/h, so the horizontal distance the target travels is ...
x = (109 - 81.0)/3.6·t + 13.4 = 70t/9 +13.4 . . . . meters in t seconds
We want the target distance to be the same as the grenade's distance, so the time required to hit the target is ...
d = x
4.9t² = (70/9)t +13.4
4.9(t² -(100/63)t) = 13.4
4.9(t -50/63)² = 13.4 +4.9(50/63)² ≈ 16.486
t = 50/63 + √(16.486/4.9) ≈ 2.62793 . . . seconds
This corresponds to a launch speed of ...
v = (4.9√2)t ≈ 18.21 . . . . meters/second
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B. As we said, 1 m/s = 3.6 km/h, so the launch speed is about 65.558 km/h at an angle of 45° relative to the direction of travel. The magnitude of the velocity relative to the earth (ve) will be this vector added to 81.0 km/h in the direction of travel. The (horizontal, vertical) components of that sum are ...
(81.0, 0) + 65.558(cos(45°), sin(45°)) ≈ (127.357, 46.357) km/h
The magnitude is the Pythagorean sum:
ve = √(127.357² +46.357²) ≈ 135.5 . . . . km/h
