The shot put is an athletics event in which competitors "throw" a shot as far as possible - Scottish Highers Physics - Question 1 - 2015

To calculate the horizontal component of the initial velocity ($v_x$), we use the formula:

$v_x = v imes ext{cos}( heta)$

Substituting in the values:

≈ 11 imes 0.766 = 8.43 ext{ms}^{-1}$$ So, the horizontal component is approximately 8.4 m/s.

To find the vertical component of the initial velocity ($v_y$), we use:

$v_y = v imes ext{sin}( heta)$

Thus:

≈ 11 imes 0.643 = 7.07 ext{ms}^{-1}$$ The vertical component is approximately 7.1 m/s.

To calculate the total time of flight ($T$), we need to consider the time to reach maximum height and the time to fall back down: $T = t_{ ext{up}} + t_{ ext{down}}$

Given that the time to reach maximum height is 0.76 s and using the symmetry of projectile motion, the time to fall back down is approximately the same: $T = 0.76 s + 0.76 s = 1.52 s.$

Thus, the total time of flight is approximately 1.52 seconds.

The range ($R$) of the shot can be calculated using the formula:

$R = v_x imes T$

Substituting the values: $R = 8.43 ext{ ms}^{-1} imes 1.52 ext{ s} ≈ 12.8 ext{ m}$

Therefore, the range of the shot is approximately 12.8 meters.

Increasing the angle of projection generally increases the vertical component of the velocity while decreasing the horizontal component. However, the total velocity at release remains constant unless additional force is applied, hence the total kinetic energy:

KE = rac{1}{2}mv^2

is largely dependent on the release speed. Therefore, while the distribution of kinetic energy shifts with a higher angle, limiting factors like air resistance and optimal angles (usually around 45° for maximum distance) must also be considered for practical applications.