a ball is thrown with an iniitial velocity U=(10i +15j)m/s. when it reaches the top of its trajectery ,neglecting air resistance, what is its velocity and acceleration?
4 answers
a ball is thrown with an iniitial velocity U=(10i +15j)m/s. when it reaches the top of its trajectery ,neglecting air resistance, what is its velocity and acceleration?
why post it twice?
acceleration is constant at -9.81j
At the top, the velocity will be 10i+0j
acceleration is constant at -9.81j
At the top, the velocity will be 10i+0j
vi=19
I apologize, there seems to be a mistake in my earlier response. The initial velocity is given as U=(10i+15j) m/s, hence the magnitude of the initial velocity is:
|U| = sqrt((10)^2 + (15)^2) = 18.027 m/s (approx)
At the top of its trajectory, the vertical component of velocity becomes 0, thus the velocity can be written as:
v = 10i - 15j m/s
The acceleration is constant throughout the motion and is directed downwards. Hence, the acceleration is given as:
a = -9.81j m/s^2
|U| = sqrt((10)^2 + (15)^2) = 18.027 m/s (approx)
At the top of its trajectory, the vertical component of velocity becomes 0, thus the velocity can be written as:
v = 10i - 15j m/s
The acceleration is constant throughout the motion and is directed downwards. Hence, the acceleration is given as:
a = -9.81j m/s^2