In a shotput event an athlete throws the shotput of mass 10 kg
with an initial speed of 1m s'at 45° from a height 1.5 m above
ground. Assuming air resistance to be negligible and acceleration
due to gravity to be 10 ms 2. the kinetic energy of the shotput
when it just reaches the ground will be
Answers
Answer:
the kinetic energy of the shotput when it just reaches the ground will be = 155 kgm/s
Explanation:
initial speed of 1m/s at 45°
Horizontal Speed = 1 Cos45 = 1/√2 m/s
Vertical Speed = 1 Sin45 = 1/√2 m/s
Vertical Speed at top = 0
Time taken to reach at top = (0 - 1/√2)/(-10) = 1/10√2 Sec
Vertical Distance Covered = (0² - (1/√2)²)/(2(-10)) = 1/40 m = 0.025 m
Max height reached = 1.5 + 0.025 m = 1.525 m
Vertical velocity when Reached at Ground
V² = 2 (10) * 1.525
=> V² = 30.5 m
KE = (1/2) * Mass ( Vhorizontal² + Vvertical²)
= (1/2) 10 * (1/2 + 30.5)
= 310/2
= 155 kgm/s
the kinetic energy of the shotput when it just reaches the ground will be = 155 kgm/s
Another Simple way to solve :
At time of throw
KE = (1/2)mV² = (1/2)(10)1² = 5 kgm/s
PE = mgh = 10 * 10 * 1.5 = 150 kgm/s
TE = 5 + 150 = 155 kgm/s
When Hit ground
PE = 0 as h = 0
so KE = TE - PE = 155 - 0 = 155 kgm/s