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?

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

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s