Question

A boy stands on the roof of a 50m building and throws a baseball at an angle of 25 degrees above the horizontal at 10m/s. How far from the foot of the building will the ball land ?

Answer 1

$We \: have \: to \: simply \: calculate \: the \: range \: of \: projectile.$

$R = \frac{ {u}^{2}sin {2}^{0} }{g}$

$R = \frac{ {12}^{2} \sin(2 \times {30}^{o} ) }{10}$

$5 \sqrt{3} = 8.66m$

Answer 2

Given:

Initial speed, u = 10 m/s

Height of building, h = 50 m

Angle, θ = 25°

Find:

The distance of the ball from the foot of the building i.e., horizontal range (R).

Solution:

The horizontal component of velocity is

$u_x = 10cos \theta = 10cos25\textdegree = (10)(0.9063) = 9.063 m/s$

The vertical component of velocity is

$u_y = 10sin \theta = 10sin25\textdegree = (10)(0.4226) = 4.226 m/s$

If the ball hits the land after t seconds, then range of projectile is given by

$R = u_x\times t$

$R = 9.063 t$

Let us now consider the vertical motion of the ball.

$h = -u_y t + \frac{1}{2} gt^2$

where h = 50 m

$u_y$ = 4.226 m/s

t = R/9.063

g = 10 m/s2

Putting values in the equation, we have

$50 = -4.226t + 5t^2$

$50 = -4.226\frac{R}{9.063} + 5(\frac{R}{9.063} )^2$

$50 = 0.4663R + 0.0609R^2\\0.0609R^2 + 0.4663R - 50 = 0$

Solving the above quadratic equation, we get

R = -32.7 m or 25.1 m

Neglecting negative value, we have

R = 25.1 m = 25 m (approx.)

Hence, The ball will land 25 m away from the foot of the building.

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