Question

A ball is thrown upward at an angle of 37⁰ to the horizontal and lands on the the top edge of a building that is 20m away. the top edge is 10m above the throwing point. how fast was the ball thrown?

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Answer 1

$\begin{gathered}\Large\bf{\color{green}GiVeN,} \\ \end{gathered}$

$\begin{gathered}\bf\pink{As\:shown\:in\:figure,} \\ \end{gathered}$

A ball is thrown upward at an angle of 37° to the horizontal.

$\begin{gathered}:\implies\:\:\bf\blue{\theta\:=\:37°} \\ \end{gathered}$

And lands on the top edge of a building, i.e. 20 m away, i.e. the horizontal component is 20 m.

$\begin{gathered}:\implies\:\:\bf\green{x\:=\:20\:m} \\ \end{gathered}$

The top edge is 10 m above the throwing point, i.e. the vertical component is 10 m.

$\begin{gathered}:\implies\:\:\bf\orange{y\:=\:10\:m} \\ \end{gathered}$

$\begin{gathered}\Large\bf{\color{lime}To\:FiNd,} \\ \end{gathered}$

How fast was the ball thrown.

$\begin{gathered}\Large\bf{\color{indigo}CaLcUlAtIoN,} \\ \end{gathered}$

☆ The key of this problem is to treat the horizontal and vertical components of the motion separately.

$\begin{gathered}\bf\purple{We\:know\:that,} \\ \end{gathered}$

$\begin{gathered}\red\bigstar\:\:\bf{\color{peru}y\:=\:x\:\tan{\theta}\:-\:\dfrac{gx^2}{2u^2\:\cos^2{\theta}}\:} \\ \end{gathered}$

$\begin{gathered}\bf\orange{Where,} \\ \end{gathered}$

y = 10 m

x = 20 m

θ = 37°

$\begin{gathered}\longmapsto\:\:\bf{\tan{37°}\:=\:\dfrac{3}{4}\:} \\ \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf{\cos{37°}\:=\:\dfrac{4}{5}\:} \\ \end{gathered}$

g = 9.8 m/s²

u = Initial velocity

$\begin{gathered}:\implies\:\:\bf{10\:=\:\Big(20\times{\tan{37°}}\Big)\:-\:\dfrac{9.8\times{(20)}^2}{2u^2\:\cos^2{37°}}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{10\:=\:\Big(20\times{\dfrac{3}{4}}\Big)\:-\:\dfrac{3920}{2u^2\times{\Big(\dfrac{4}{5}}\Big)^2}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{\dfrac{3920}{2u^2\times{\dfrac{16}{25}}}\:=\:15\:-\:10\:} \\ \end{gathered}$

$</p><p>\begin{gathered}:\implies\:\:\bf{\dfrac{3920}{2u^2}\times {\dfrac{25}{16}}\:=\:15\:-\:10\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{\dfrac{245\times{25}}{2u^2}\:=\:5\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{2u^2\:=\:\dfrac{245\times{25}}{5}\:} \\ \end{gathered}$

$</p><p>\begin{gathered}:\implies\:\:\bf{2u^2\:=\:1225\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{u^2\:=\:\dfrac{1225}{2}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{\color{olive}u\:=\:24.74\:m/s\:} \\ \end{gathered}$

$\Large\bold{therefore}$

∴ The ball was thrown at 24.74 m/s.

$\boxed{\huge \star{{\mathfrak{\red{☠Naitik \: : hêrë☠}\star}}}}$

Answer 2

Answer

Ball was thrown with a velocity of about 24.75 m/s

Explanation

Given:-

The angle of projection of the ball, θ = 37°

Distance of the building from the point of projection, x = 20 m

Height of the building, y = 10 m

Ball lands on the top edge of the building, so position of the ball in terms of coordinates = (20, 10)

To find:-

Velocity of the ball when thrown, u =?

Formula required:-

Trajectory formula for projectile motion

        y = x tan θ - gx²/(2 u² cos²θ)

where

y = vertical position of body

x = horizontal position of body

θ = angle of projection

g = acceleration due to gravity

u = Initial velocity of the body while projecting

Solution:-

Using the trajectory formula

→ y = x tan θ - gx²/(2 u² cos²θ)

[ Putting known values ]

→ 10 = 20 × tan 37° - g × (20)² / (2 u² cos²37°)

[ taking, g = 9.8 m/s²; tan 37° = 3/4 and cos 37° = 4/5 ]

→ 10 = (20 × 3/4) - (9.8 × 400) / (2 × u² × (4/5)²)

→ 10 = (15) - 3920/(32 u²/25)

→ 10 - 15 = -98000/(32 u²)

→ -5 = -98000/(32 u²)

→ 160 u² = 98000

→ u² = 612.5

→ u ≈ 24.75 m/s

Therefore,

The Initial velocity of the ball when thrown is 24.75 m/s.