A cannonball, m= 5 kg, is shot out of a cannon at an initial speed of 5 m/s, at

Question

A cannonball, m= 5 kg, is shot out of a cannon at an initial speed of 5 m/s, at an angle theta = 25 degree with respect to the horizontal. The cannonball leaves the cannon at an initial height of 0.6m with respect to the ground. Define +i to the right, and +j as up. Find the velocity of the cannonball at the maximum height. NOTE: It is NOT zero! Also, be sure to include the DIRECTION. Find the time it takes to reach the maximum height. Find the velocity of the cannonball just before it hits the ground. Include the DIRECTION. Two masses are connected by a thin string, which is across a frictionless pulley. The mass on the ramp, mr = 10kg, while the hanging mass mh = 8.0kg. There is no friction between the ramp and the mass. Obtain the acceleration of the system. In terms of direction, state whether the acceleration of the hanging mass is upwards or downwards. Obtain the tension T in the string. Leaving the ramp mass alone, for what value of the hanging mass would you need so that the acceleration is zero? A rifle is aimed horizontally at a target 26 m away. The bullet hits the target 1.8 times 10 -2 m below the aiming point. Neglect air resistance. Define up as + and towards the target as + What is the bullet's time of flight? What is the bullets velocity just before impact? Suppose I wanted to hit the target at a height equal to the initial height of the gun. At what angle do I aim the gun?

Explanation / Answer

this is a case of projectile motion

at max height the vertical component of velocity is zero

but horizontal conpoment remains constant

1)so velocity at max height = Vcos = 5cos25 = 4.53 m/s

direction = +i

2)time of flight = 2vsin/g

time taken to attain max height = vsin/g = .2156 s

3)velocity on reaching the ground = vcos25 i + (.5*9.8*(.2156)^2 + 2*9.8*.6)

= (4.513 i + 3.6569 -j ) m/s

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