Question

Problems
1. A girl stands on the roof of 55m tall building and throws a baseball
at an angle of 30° above horizontal at an angle of 10m/s. (4 Marks)
a. How long the ball is in air.
b. How far from the foot of the building the ball land?

Answer 1

initial velocity of ball, u = 10m/s at an angle 30° with horizontal.

so, vertical component of initial velocity of ball, $u_y=usin\theta$

= 10sin30° = 5 m/s

vertical position of ball, y = 55m

using formula,

$y=u_yt+\frac{1}{2}a_yt^2$

taking reference level is top of the building.

then, y = -55 , $u_y=5m/s$

a_y =-g = -10m/s²

so, -55 = 5t + 1/2 (-10)t²

⇒-11 = t - t²

⇒t² - t - 11 = 0

⇒t = {1 ± √(1 + 44)}/2

t = 3.8541 sec

now , horizontal component of initial velocity of ball, $u_x=10cos\theta$

= 10cos30° = 5√3 m/s

now using formula, $x=u_xt+\frac{1}{2}a_xt^2$

here, $a_x=0$

so, x = 5√3 × 3.8541 ≈ 33.377 m

Answer 2

$\huge\bold\purple{Answer:-}$

y = -55 , u_y=5m/s

a_y =-g = -10m/s²

so, -55 = 5t + 1/2 (-10)t²

⇒-11 = t - t²

⇒t² - t - 11 = 0

⇒t = {1 ± √(1 + 44)}/2

t = 3.8541 sec

now , horizontal component of initial velocity of ball, u_x=10cos\theta

= 10cos30° = 5√3 m/s

now using formula, x=u_xt+\frac{1}{2}a_xt^2

here, a_x=0

so, x = 5√3 × 3.8541 ≈ 33.377 m