A body is thrown with a velocity of 10m/s at an angle of 60 degrees wi...
At the highest point, its vertical velocity will become zero. Hence the only left velocity is horizontal which remains unchanged and is equal to 10 cos 60 = 5m/s
A body is thrown with a velocity of 10m/s at an angle of 60 degrees wi...
Given, initial velocity, u = 10m/s and angle of projection, θ = 60 degrees
At highest point, vertical component of velocity, v = 0
To find: Velocity at highest point, Vh
We can use the formula for vertical component of velocity, v = u sin θ - gt
Where g is the acceleration due to gravity and t is the time taken to reach the highest point.
Since v = 0 at highest point, we can solve for t:
0 = u sin θ - gt
t = u sin θ/g
Substituting the values, we get:
t = (10 sin 60)/9.8
t = 1.02 s
Now, we can use the formula for velocity in the vertical direction at any time t, given by:
v = u sin θ - gt
At the highest point, v = 0, so we have:
0 = 10 sin 60 - 9.8(1.02)
0 = 5 - 9.996
-4.996 = -5 (approx.)
Therefore, the velocity at the highest point is approximately 5 m/s.
Hence, the correct answer is option 'A' (5 m/s).