CAIE M2 2013 June — Question 3

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2013
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeHorizontal projection from height
DifficultyModerate -0.8 This is a straightforward projectile motion problem requiring standard SUVAT equations in two dimensions. Students apply horizontal and vertical motion independently, then combine components using Pythagoras and trigonometry. All steps are routine with no problem-solving insight required, making it easier than average but not trivial due to the two-part calculation.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model

3 A ball is projected horizontally with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from the top of a tower which is 30 m high. The tower stands on horizontal ground.
  1. Find the speed and direction of motion of the ball when it reaches the ground.
  2. Calculate the distance from the foot of the tower to the point where the ball reaches the ground.

3 A ball is projected horizontally with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the top of a tower which is 30 m high. The tower stands on horizontal ground.\\
(i) Find the speed and direction of motion of the ball when it reaches the ground.\\
(ii) Calculate the distance from the foot of the tower to the point where the ball reaches the ground.

\hfill \mbox{\textit{CAIE M2 2013 Q3}}