Answer:
v₀ = 16.55 m/s
Explanation:
This motion of the ball can be modeled as a projectile motion with following data:
R = Range of Projectile = 27.5 m
θ = Launch Angle = 50°
g = acceleration due to gravity = 9.81 m/s²
v₀ = Initial Speed of Ball = ?
Therefore, using formula for range of projectile, we have:
<u>v₀ = 16.55 m/s</u>
What’s the question here?
B. Both of these types of lenses have the ability to produce real images.
Answer:
(A) 11 m/s
(B) 1.3 m
Explanation:
Horizontal range, R = 9.6 m
Angle of projection, theta = 28 degree
(A)
Use the formula of horizontal range
R = u^2 Sin 2 theta / g
u^2 = R g / Sin 2 theta
u^2 = 9.6 × 9.8 / Sin ( 2 × 28)
u = 10.65 m/s
u = 11 m/s
(B)
Use the formula for maximum height
H = u^2 Sin ^2 theta / 2g
H =
10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)
H = 1.275 m
H = 1 .3 m
