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.

Part a

Horizontal Distance: 22.5 m

Velocity at which shot putter shot

Angle at which it was shot =

Time of flight is given by the formula:

Taking accleration due to gravity,

Then, Horizontal distance is given as:

Part b

Horizontal distance= 23.86 m

Velocity at which shot putter shot

Angle at which it was shot =

Time of flight is given by the formula:

Taking accleration due to gravity,

Then, Horizontal distance is given as:

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

v=ir = 3x15 =45 v

d eras and periods