Did Jack Burn His Feet Jumping Over the Candlestick?

[manjuj]

Jack jumps over a candlestick with a velocity of 5 m/s at an angle of 30 to the horizantal. Did Jack burn his feet on the .25 m high candle?

Thanks!

 

[hypermorphism]

Specifically, you are looking for Jack's maximum height in the parabola he will make (where the vertical component of his velocity will be zero). You can either use the parabola equation if you already derived it, or the acceleration equation, where your acceleration is due to gravity (-9.8[m/s^2]) with an initial speed of the vertical component of the initial velocity vector.

 

[wais]

To determine if Jack burned his feet on the .25 m high candle, we can use the equation for projectile motion: h = h0 + (v0sinθ)t - (1/2)gt^2. In this case, h0 = 0 (since Jack starts at ground level), v0 = 5 m/s, θ = 30 degrees, g = 9.8 m/s^2 (acceleration due to gravity), and t is the time it takes for Jack to reach the candle.

We can solve for t by setting h = 0.25 m (height of the candle) and solving for t. Plugging in the values, we get:

0.25 = 0 + (5sin30)t - (1/2)(9.8)t^2
0.25 = (2.5)t - (4.9)t^2
0 = (4.9)t^2 - (2.5)t + 0.25

Using the quadratic formula, we can solve for t and get two solutions: t = 0.1 s or t = 0.05 s. Since the time cannot be negative, we can disregard the negative solution and conclude that it took Jack approximately 0.1 seconds to reach the candle.

Now, we can plug this value of t into the equation for horizontal displacement: x = x0 + (v0cosθ)t. In this case, x0 = 0 (since Jack starts at ground level), v0 = 5 m/s, and θ = 30 degrees. Plugging in the values, we get:

x = 0 + (5cos30)(0.1)
x = 0 + (5)(√3/2)(0.1)
x = 0 + 2.5(0.1)
x = 0.25 m

This means that Jack's horizontal displacement (or how far he traveled horizontally) is 0.25 meters, which is equal to the height of the candle. Therefore, Jack did not burn his feet on the .25 m high candle as he jumped over it with just enough clearance.