Question

A batter hits a baseball so that it leaves the bat at a speed of 32 m/s at an angle 51 degrees to the horizontal. How far (in meters) from where it was struck will it land? Assume no air resistance and g=-9.8 m/s2 2DP?

A motorcycle stunt rider rides off the edge of a 500 m cliff. Just at the edge, his velocity is horizontal, with magnitude 16 m/s. What is the magnitude of the motorcycle's velocity (in m/s) after 5.6 s? Assume no air resistance, and use g=-9.8 m/s2. 2 DP

Answer 1

1)

Use formula $R=u^{2}sin2\theta /g$

$R=(32m/s)^{2}sin2*51 /(9.81m/s^{2})$

$R=32^{2}sin102 /9.81$

ANSWER: ${\color{Red} R=102.1m}$

=================================

2)

Since there is no acceleration in horizontal direction ,horizontal velocity will be same

Consider vertical motion

initial vertical velocity = zero

Use formula $v=u+at$

$v_{y}=u_{y}+gt$

$v_{y}=0m/s+(9.81m/s^{2})*5.6s$

$v_{y}=0+9.81*5.6$

${\color{Red} v_{y}=54.936m/s}$

${\color{Red} v_{x}=16m/s}$

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{(16m/s)^{2}+(54.936m/s)^{2}}$

ANSWER: ${\color{Red} v=57.22m/s}$

========================