IB Physics SL A 1 8 Projectile Components Questions

Practise IB Physics SL A.1.8 by resolving projectile velocity into horizontal and vertical components and applying independent motion equations.

Exam points

Resolve an initial projectile velocity into horizontal and vertical components using the stated launch angle and direction.
Treat the horizontal and vertical motions independently when air resistance is absent, identifying which component changes under gravity and why objects with different horizontal speeds can share a fall time.
Use the component equations to calculate time of flight, horizontal range, maximum height, impact speed or the position at a boundary such as a wall, net or cliff edge.
Interpret or sketch the component velocity and acceleration graphs, including signs, the highest point and the continuing downward acceleration there.
Compare projectile outcomes when launch speed, angle, height or gravitational field changes, using the relevant component and stated constraint rather than assuming the standard same-height formula.

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Study A.1.8—Projectile components conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 5]

A student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of $64 \mathrm{~m} \mathrm{~s}^{-1}$ . The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms .

(a)

The student strikes the tennis ball at point P . The tennis ball is initially directed at an angle of $7.00^{\circ}$ to the horizontal.

The following data are available.

[ 5 ]

(i)

Calculate the time it takes the tennis ball to reach the net.

[ 2 ]

horizontal component of velocity is $64.0 \times \cos 7^{\circ}=63.52<\mathrm{ms}^{-1} \gg$ $t=<\frac{11.9}{63.52}=>0.187 / 0.19<\mathrm{s} \gg$

Marking guidance:

Do not award BCA. Check working. Do not award ECF from using $64 \mathrm{~m} \mathrm{~s}^{-1}$ .

(ii)

Show that the tennis ball passes over the net.

[ 3 ]

ALTERNATIVE 1
$u_{y}=64 \sin 7 / 7.80<\mathrm{ms}^{-1} \gg$

decrease in height $=7.80 \times 0.187+\frac{1}{2} \times 9.81 \times 0.187^{2} / 1.63$ «m»

final height = « 2.80-1.63 » =1.1 / 1.2 «m»

«higher than net so goes over»

ALTERNATIVE 2

vertical distance to fall to net «=2.80-0.91 »=1.89 «m»

time to fall this distance found using $<1.89=7.8 t+\frac{1}{2} \times 9.81 \times t^{2}$ »

t=0.21 «s»

0.21 «s» > 0.187 «s»

«reaches the net before it has fallen far enough so goes over»

Other alternatives are possible

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IB Physics SL A 1 8 Projectile Components Questions question 1

Question (a)

Question (i)

Question (ii)