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In 1978, Geoff Capes of the United Kingdom won a competition for throwing 5 lb bricks; he threw one brick a distance of 44.0 m. Suppose the brick left Capes' hand at an angle of 45.0° with respect to the horizontal a. What was the initial speed of the brick? b. What was the maximum height reached by the brick?

Respuesta :

Answer:

A) 20.8 m/s

B) h_max = 11 m

Explanation:

A) Formula for projectile range is;

R = (u²sin2θ)/g

We want to find initial velocity, so let's make u the subject.

u = Rg/sin2θ

We are given;

R = 44 m

θ = 45°

Thus;

u = √[(44 × 9.8)/sin 2(45)]

u = √[431.2/sin 90]

u = 20.77 m/s ≈ 20.8 m/s

B) maximum height will be gotten from the formula;

h_max = (R tan θ)/4

h_max = (44 × tan 45)/4

h_max = (44 × 1)/4

h_max = 11 m

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