A ball is thrown straight up with a velocity of 50 m/s.(use g = -10 m/s^2) a) What is the velocity in m/s after 2 seconds?
1 answer:
Answer:
(a) 30 m/sec
(b) -50 m/sec
Explanation:
We have given initial velocity of ball u = 50 m/sec
Acceleration due to gravity
(a) Time t = 2 sec
Now according to first equation of v = u-gt
So v=50-10×2=30 m/sec
(b) Time t = 10 sec
Now according to first equation of motion
So final velocity v = u-gt = 50-10×10 =-50 m/sec
You might be interested in
Explanation:
Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.
→ The tangential component of acceleration is rate of increase in the speed of plane so,
→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).
dy/dx = d(0.4x²)/dx
= 0.8x
Take the derivative again,
d²y/dx² = d(0.8x)/dx
= 0.8
at x= 5 Km
dy/dx = 0.8(5)
= 4
now insert the values,
→ Now the normal component of acceleration is given by
= (200)²/(87.6×10³)
aₙ = 0.457 m/s²
→ Now the total acceleration is,
a = 0.921 m/s²