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?
Answers
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.
hope it helps you ❤️...
