Answer:
A) 1.53s
B) 19.8m
C) 2.869m
Explanation:
A) The time of flight for a projectile can be calculated using the formula:
t = 2μsinθ/g
Where; u = velocity
θ = angle
g = acceleration due to gravity (9.8m/s^2)
t = 2 × 15 × sin 30°/9.8
t = 30sin30°/9.8
t = 30 × 0.5/9.8
t = 15/9.8
t = 1.53s
B) The horizontal range (distance) for a projectile can be calculated using the formula:
Range = u²sin2θ/ g
Range = 15² sin 2 × 30 / 9.8
Range = 225 sin 60/9.8
Range = 225 × 0.8660/9.8
Range = 194.855/9.8
Range = 19.8m
C) The maximum height for a projectile can be calculated using the formula:
h = u²sin²θ/2g
h = 15² (sin 30)² / 2 × 9.8
h = 225 × 0.25 / 19.6
h = 56.25/19.6
h = 2.869m
Answer:
The answers are:
T, T, F, F, T, T, F, F, T, T, F
Good luck!
Answer:
V1 =8.1 m/s
Explanation:
height at highest point (h2) = 4.1 m
height at lowest point (h1) = 0.8 m
acceleration due to gravity (g) = 9.8 m/s^{2}
from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore
mgh1 +
= mgh2 + 
where
- speed at highest point = V2
- speed at lowest point = V1
- mass of the girl and swing = m
- at the highest point, the speed is minimum (V1 = 0)
- at the lowest point the speed is maximum (V2 is the maximum speed)
- therefore the equation becomes mgh1 +
= mgh2
m(gh1 +
) = m(gh2)
gh1 +
= gh2
V1 = 
now we can substitute all required values into the equation above.
V1 = 
V1 = 
V1 =8.1 m/s
Ball is thrown at an angle of 20 degree with velocity 18 m/s
so first we will find the two components of the velocity


similarly for other component


now the ball will cover horizontal distance of 30 m till it reach to other player
so here we can use the formula of time




so it will take 1.77 s to reach the other player