IB DP Physics 2.1 – Motion: IB Style Question Bank HL Paper 1


A boy throws a ball horizontally at a speed of 15 m s1 from the top of a cliff that is 80 m above the surface of the sea. Air resistance is negligible.

What is the distance from the bottom of the cliff to the point where the ball lands in the sea?

    1. 45 m

    2. 60 m

    3. 80 m

    4. 240 m


Ans: B

Assuming constant vertical acceleration, no horizontal acceleration, and no air resistance:

\(y(t) =y_0+\dot{y_0}t+\frac{1}{2}\ddot{y}t^2\)

Then when \(y(t_1)=y_1\)


so by quadratic formula

\(t_1=\frac{-\dot{y_0}\pm \sqrt{\dot{y}_0^2-4 \times \frac{\ddot{y}}{2}\times(y_0-y_1)}}{2 \times \frac{\ddot{y}}{2}}\)


\(\ddot{y}=g< 0\)
\(t_1\geq 0\)

we can reduce that to


In that same time the stone has moved \(\dot{x_0}t_1 \;m\) horizontally.


\(\dot{x_0}=15 ms^{-1}\)
\(y_0=80m \)
\(y_1=0 m\)
\(g=-9.81 ms^{-2}\)

we get

\(x_1\) =  60 m


A ball is thrown from point X and follows path XYZ. Air resistance is negligible.

Which quantity is zero when the ball is at the highest point Y of the path?

A. The horizontal component of the ball’s acceleration

B. The vertical component of the ball’s acceleration

C. The horizontal component of the ball’s velocity

D. The kinetic energy of the ball




Normally in problems on projectile motion posed to students, the only acceleration on the body is the acceleration due to gravity, whose direction is vertically downward.

So, if we have a Cartesian coordinate system with the positive X axis in the forward direction of the motion of the projectile and the positive Y axis in the upward direction, the horizontal component of the acceleration would be \(0\) and the vertical component of the acceleration would be \(-g\) during both, the upward and downward path of the projectile.

If we have a more difficult problem in which we are considering wind resistance, then the wind resistance would result in an acceleration throughout the flight of the projectile opposing the motion of the projectile.

In the vertical direction, the wind resistance would result in a negative acceleration (i.e. in the downward direction) during the upward motion and a positive acceleration (i.e. in the upward direction) during the downward motion.

The acceleration due to wind resistance is not constant and is dependent on the velocity of the projectile.

In case the projectile is magnetic, there would be an acceleration caused by the magnetic field of the earth. This would, however, be negligible compared to the other accelerations on the projectile.


Two isolated spherical planets have the same gravitational potential at their surfaces. Which ratio must also be the same for the two planets?

A. \(\frac{{{\rm{radiu}}{{\rm{s}}^{\rm{3}}}}}{{{\rm{mass}}}}\)

B. \(\frac{{{\rm{radiu}}{{\rm{s}}^{\rm{2}}}}}{{{\rm{mass}}}}\)

C. \(\frac{{{\rm{radius}}}}{{{\rm{mass}}}}\)

D. radius




The amount of work done in moving a unit test mass from infinity into the gravitational influence of source mass is known as gravitational potential.

Simply, it is the gravitational potential energy possessed by a unit test mass

⇒ V = U/m

⇒ V = -GM/R

Now two planet has same potential on surface means Vp1=Vp2




A ball of mass m is projected horizontally with speed v from a height h above the floor. Air resistance is negligible.

The horizontal distance travelled by the ball to the point where it lands on the floor depends on

A. m and h only.
B. m and v only.
C. h and v only.
D. mh and v.




Refer : https://www.iitianacademy.com/ib-physics-unit-2-mechanics-motion-notes/

If h is the distance of the ground from the point of projection, T is the time taken to strike the ground and R is the horizontal range of the projectile then
T = and
Case 1 : If the projectile is projected from the top of the tower of height ‘h’, in horizontal direction, then the height of tower h, range x and time of flight t are related as :


A student throws a stone with velocity \(v\) at an angle \(\theta \) to the vertical from the surface of a lake. Air resistance can be ignored. The acceleration due to gravity is \(g\).

What is the time taken for the stone to hit the surface of the lake?

A.     \(\frac{{v\sin \theta }}{g}\)

B.     \(\frac{{v\cos \theta }}{g}\)

C.     \(\frac{{2v\sin \theta }}{g}\)

D.     \(\frac{{2v\cos \theta }}{g}\)




The time of flight of the projectile is given by
where ‘t’ is the time of ascent or descent.
If projection angle is  from vertical then, angle of projection is  .
\(T=2t=\frac{2usin(90-\theta )}{g}=\frac{2ucos\theta }{g}\)