Question

Suppose that a shot putter can put a shot at the worldclass speed v0 = 15.00 m/s and at a height of 2.160 m. what horizontal distance would the shot travel if the launch angle θ0 is (a) 45.00° and (b) 42.00°? the answers indicate that the angle of 45°, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

Answer 1

Part a

Answer:

Horizontal Distance: 22.5 m

Velocity at which shot putter shot $v_o= 15 m/s$

Angle at which it was shot = $45^o$

Time of flight is given by the formula:

$t=\frac{2v_osin\Theta}{g}$

Taking accleration due to gravity, $g=10/s^2$

$t=\frac{2\times sin45^o}{10}=2.12 s$

Then, Horizontal distance is given as:

$x=v_x t$

$\Rightarrow x=v_ocos\Theta t$

$\Rightarrow x= 15\times cos45^o\times2.12 s=22.5 m$

Part b

Answer: Horizontal distance= 23.86 m

Velocity at which shot putter shot $v_o= 15 m/s$

Angle at which it was shot = $42^o$

Time of flight is given by the formula:

$t=\frac{2v_osin\Theta}{g}$

Taking accleration due to gravity, $g=10/s^2$

$t=\frac{2\times sin42^o}{10}=2.01s$

Then, Horizontal distance is given as:

$x=v_x t$

$\Rightarrow x=v_ocos\Theta t$

$\Rightarrow x= 15\times cos42^o\times2.01 s=23.86m$

Therefore, although the range is maximum for part a but horizontal distance covered is smaller than in part b.