On planet X, an object dropped from a height of 20 m reaches the ground in 16 seconds. What should be the magnitude of its velocity when it is launched at 60 degrees so that the range is 120 m.
On planet X, an object dropped from a height of 20 m reaches the ground in 16 seconds. To proceed the acceleration due to gravity on the planet needs to be determined. This can be done by substituting the values given in the relation s = u*t + (1/2)*a*t^2, where s is the distance traveled in time t, u is the initial velocity and a is the acceleration . This gives 20 = 0*t + (1/2)*a*16^2
a = (5/64) m/s^2.
When an object is launched at a velocity v at an angle x, the range of the object is given by R = v^2*sin 2x/g. Here, g = 5/64, x = 60 and R = 120
120 = v^2*sin 120*64/5
=> v^2 = 120*5/(sin 120*64)
=> v^2 = 21.65
=> v = 4.653
If the object is launched with a velocity of magnitude 4.653 m/s at an angle 60 degrees, it has a range of 120 m.