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=(2v_osin\Theta)/(g)`

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

`t=(2* 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* cos45^o*2.12 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=(2v_osin\Theta)/(g)`

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

`t=(2* sin42^o)/(10)=2.01s`

Then, Horizontal distance is given as:

`x=v_x t`

`\Rightarrow x=v_ocos\Theta t`

`\Rightarrow x= 15* cos42^o*2.01 s=23.86m`

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