Vikas TU
Last Activity: 4 Years ago
Given in the question ,
Height h = 40 m.
Initial velocity u = 50 m/s.
Now potential of projectile w.r.t the ground at the time it projected,
P.E= mgh
P.E= mg × 40 J
K.E when projected = 1/2 mv²
= m × 1250 J.
Now total energy at time of the projection ,
Kinetic Energy + Potential Energy
(mg × 40) + ( m × 1250) J.
assume that the speed of the projectile when it comes to ground = V
Then Kinetic energy = 1/2 mV² J
So Total energy = K.E + P.E = 1/2 mV²
1/2 mV² = (mg × 40 ) + (m × 1250 ) J
V² = 80 × g + 2500
V² = 80 × 9.8 + 2500
V²= 3284
V =√3284
V= 57.30 m/s
Hence the speed when it hit ground is 57.30 m/s.
