Question : A ball is thrown from the top of a tower with an initial velocity of 10 m/s at an angle of 30° with the horizontal. If it hits the ground at a distance of 17.3 m from the base of the tower, calculate the height of the tower. Given g = 10 m/s².
Doubt by Girisha
Solution :
Here
u=10 m/s
θ=30°
R=17.3 m
Considering the horizontal motion of the projectile
R=Speed×time
R=ucosθ×t
17.3=10×cos30°×t
17.3=10×(√3/2)×t
17.3 = 5√3×t
t=17.3/5√3
t=17.3/(5×1.73)
t=17.3/8.65
t =17.3/8.66 = 2 s
Now, consider the vertical motion.
Initial vertical velocity,
uy=usin30°
=10×(1/2)
= 5 m/s
Using the equation,
s=ut+1/2at²
Taking upward as positive,
-h=5×2+1/2×(-10)×(2)²
-h = 10 - 20
-h = -10
h = 10 m
Hence, the height of the tower is 10 m.