Kinematics of a particle moving in a straight line.
A/S Physics
Take g=9.81ms-2
Answers
Qu 1.A particle is projected horizontally at 15ms-1 from a point 60m above a horizontal surface. Find how long it takes before the particle strikes the surface and the horizontal distance travelled in that time.
Qu 2.A particle is projected horizontally at 20ms-1. It strikes a horizontal surface after travelling 50m horizontally. Find the height of the point of projection above the surface.
Qu3A stone is thrown horizontally from a window 3.6m above horizontal ground. It hits the ground after travelling 10m horizontally. Find the speed of projection.
Qu 4.A particle is projected with a speed of 49ms-1 at an angle of 45 degrees above the horizontal. Find (a) the time taken by the particle to reach its maximum height (b) the maximum height reached (c) the time of flight (d) the horizontal range of the particle.
Qu 5A particle is projected with a speed of 56ms-1 at an angle of 30 degrees above the horizontal. Find (a) the time taken by the particle to reach its maximum height (b) the maximum height reached (c) the time of flight (d) the horizontal range of the particle.
Qu 6.A particle is projected with a speed of 28ms-1 at an angle above the horizontal. If the greatest height reached above the point of projection is 14m find the value of .
Kinematics of a particle moving in a straight line.
A/S Physics
Take g=9.81ms-2
Qu 1.A particle is projected horizontally at 15ms-1 from a point 60m above a horizontal surface. Find how long it takes before the particle strikes the surface and the horizontal distance travelled in that time.
Answer:
Using:
Rearranging and substituting:
Now the horizontal distance travelled is just m
Qu 2.A particle is projected horizontally at 20ms-1. It strikes a horizontal surface after travelling 50m horizontally. Find the height of the point of projection above the surface.
Answer:
So we can now work out how long it was in flight:
And then:
Qu3A stone is thrown horizontally from a window 3.6m above horizontal ground. It hits the ground after travelling 10m horizontally. Find the speed of projection.
Answer:
Time in flight is just
Hence the speed is just:
ms-1
Qu 4.A particle is projected with a speed of 49ms-1 at an angle of 45 degrees above the horizontal. Find (a) the time taken by the particle to reach its maximum height (b) the maximum height reached (c) the time of flight (d) the horizontal range of the particle.
Answer:
(a)Calculate the vertical component of the initial velocity first:
When it is at its maximum height, the vertical velocity is zero, so:
(b)The maximum height is reached when :
(c)The time of flight is just twice the time it takes to reach maximum height (by symmetry), 7.06s.
(d)The horizontal component of velocity does not change, so its just speed times time in flight:
Qu 5A particle is projected with a speed of 56ms-1 at an angle of 30 degrees above the horizontal. Find (a) the time taken by the particle to reach its maximum height (b) the maximum height reached (c) the time of flight (d) the horizontal range of the particle.
Answer
(a)Calculate the vertical component of the initial velocity first:
When it is at its maximum height, the vertical velocity is zero, so:
(b)The maximum height is reached when :
(c)The time of flight is just twice the time it takes to reach maximum height (by symmetry), 5.71s.
(d)The horizontal component of velocity does not change, so its just speed times time in flight:
Qu 6.A particle is projected with a speed of 28ms-1 at an angle above the horizontal. If the greatest height reached above the point of projection is 14m find the value of .
Answer:
Just resolve the vertical component: