Question

. A golf ball is hit upward at an angle θ from the horizontal and speed v = 40 m/s. It reaches a maximum height of 10 m as illustrated above. Assume the ballistic trajectory starts at ground level and ignore air resistance. Assume the shot is made on a wide, level field.
What is the initial angle θ ?

A) θ = 20.5°
B) θ= 46.4°
C) θ = 57.3°
D) θ = 70.2°
E) θ = 81.1°
31. How long is the ball in the air?
A) 1.15 s B) 2.86 s C) 3.12 s D) 3.37 s E) 3.59 s

Answer 1

first divide the velocity of the ball in two components..vertically and horizontally...
there is a downward acceleration due to the pull of gravity..
and u know the maximum height of the ball..its 10 m
use..
\[v ^{2}_{y}=u ^{2}_{y}-2gH\]
where \(v_{y}\)=final vertical velocity
\(u_{y}\)=initial vertical velocity
H=maximum height
u can get the angle \(\theta\) from here...

Answer 2

and for the second question..use this...\[v _{y}=u _{y}-g T/2\]
where T is the time of flight

Answer 3

in the final position \(v_{y}\)=0
and at initial position \(u_{y}\)=\(u sin\theta\)

Answer 4

can you explain how to get thetha from the first equation?? @souvik

Answer 5

here the vertical component of the velocity\(u_{y}\)=\(u sin\theta\)