A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off toge
1 answer:
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When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
Answer:
<em>The water hits the wall at a height of 5.38 m</em>
Explanation:
<u>Projectile Motion </u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
The object describes a parabolic path given by the equation:
Where:
y = vertical displacement
x = horizontal displacement
θ = Elevation angle
vo = Initial speed
The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.
The height at which the water hits a wall located at x=8 m from the hose is:
Calculating:
y = 5.38 m
The water hits the wall at a height of 5.38 m