Question

A golfer standing on level ground hits a ball with
a velocity of u = 52 m/s at an angle a above the
horizontal. If tan a = 5/12, then the time interval
for which the ball is at least 15 m above the
ground (i.e. between A and B will be : (take g =
10 m/s)
15 m
15 m
(1) 1 sec
(2) 2 sec
(3) 3 sec
(4) 4 sec​

Answer 1

Answer:

(2) 2 s

Explanation:

1) We have,

$tan(a) = \frac{5}{12} \\ \\=> \frac{sin(a)}{cos(a)}=\frac{5}{12}\\ \\=>cos(a) = \frac{12}{5}sin(a)\\ \\But,\\ \\ sin^2(a)+cos^2(a)=1\\ \\=>sin^2(a)+\frac{12^2}{5^2}sin^2(a)=1\\ \\ =>sin^2(a)[1+\frac{12^2}{5^2}]=1\\ \\=>sin^2(a)=\frac{5^2}{13^2}\\ \\=>sin(a)=\frac{5}{13}$

2) Also,

$S_y=15 m\\ \\ u_y = u sin(a) = 52*\frac{5}{13} = 20 m/s \\ \\ g = 10 m/s^2$

Using Newton's Equation of motion,

$S_y=u_yt-\frac{1}{2}gt^2\\ \\=> 15 = 20t-5t^2\\ \\=>t^2-4t+3=0\\ \\=>(t-3)(t-1)=0\\ \\=>t=3,1$

3)Hence, Ball was at 15 m from ground at t= 3 s and t=1 s.

=> From t=1 to t=3 , ball was above 15 m.

=> Time Interval = 3-1=2 s

That is , Option (2) 2 s is correct answer