A projectile of mass 20.2kg is fired at an angle of 61.0(degree) above the horiz

Question

A projectile of mass 20.2kg is fired at an angle of 61.0(degree) above the horizontal and with a speed of 75.0m/s . At the highest point of its trajectory the projectile explodes into two fragments with equal mass. The first fragment falls straight down after the explosion with an initial velocity of zero in the y-direction. You can ignore air resistance.
How far from the launch point does the second fragment strike the ground if the terrain is level?

How much energy is released during the explosion?
Please Help

Explanation / Answer

first work out how high the projectile goes. You can use conservation of energy for that - Kinetic energy gets turned into Potential energy so that 1/2 mv^2 = mgh, or h = v^2/2g where v is the vertical velocity... that's 75 * sin(61 deg) m/s Next you need to think about conservation of momentum. If the projectile splits in half and one piece falls straight down, presumably the other piece obtains all it's horizontal momentum. because the weights are equal, the horizontal velocity will double. ok - the initial horizontal velocity was 75*cos(61) which works until up to the apex - after that the horizontal velocity doubles to 150*cos(61). The last bit - how long does if fly for. on the way up, v=u + at => 0 = 75sin(61) - 9.81*t. t=75*sin(61)/9.8=6.69151463 secs to go up it travels horizontal distance while going up is 75*cos(61)*t horizontal distance while going down is 150*cos(61)*t total distance is 225*cos(61)*t=730.635524 m

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