48 kg projectile is fired at an angle of 30 degree above the horizontal with an initial speed of 1.34 x10^2m/s fro the top of a cliff 150m above level ground, where the ground is taken to be y=0 what is the inital total mechanical energy of the projectile?'
To find the initial total mechanical energy of the projectile, we need to consider both the kinetic energy (KE) and potential energy (PE) at the initial position.
The kinetic energy of an object is given by the formula KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.
The potential energy of an object at a certain height above the ground is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height above the ground.
Given:
Mass of the projectile (m) = 48 kg
Initial speed (v) = 1.34 × 10^2 m/s
Height above the ground (h) = 150 m
Acceleration due to gravity (g) = 9.8 m/s^2
First, let's calculate the kinetic energy:
KE = (1/2)mv^2
= (1/2)(48 kg)(1.34 × 10^2 m/s)^2
= (1/2)(48 kg)(17956 m^2/s^2)
= 431712 J
Next, let's calculate the potential energy:
PE = mgh
= (48 kg)(9.8 m/s^2)(150 m)
= 70560 J
Finally, to find the initial total mechanical energy, we sum up the kinetic and potential energy:
Total mechanical energy = KE + PE
= 431712 J + 70560 J
= 502272 J
Therefore, the initial total mechanical energy of the projectile is 502272 Joules.