Force = mass * acceleration
15 = 2.1 * a
a = 15/2.1 = 7.143 m/s^2
The observation point on Earth and the two stars form a triangle. The two sides of the triangle are 23.3 ly and 34.76 ly and their included angle is 76.04°. We can use the cos rule to find the third side, which is the distance between the two stars.
c² = a² + b² - 2abCos(C)
c² = (23.3)² + (34.76)² - 2(23.3)(34.76)Cos(76.04)
c = 36.88 light years.
Answer:
The horizontal distance traveled by the projectile is 15.23 m.
Explanation:
Given;
angle of projection, θ = 25⁰
initial velocity of the projectile, u = 15 m/s
time of flight, t = 1.12 s
The the travelling path of the object is calculated as the range of the projectile

Therefore, the horizontal distance traveled by the projectile is 15.23 m.
Answer:
Vy = 26 m/s sin 30 = 13 m/s vertical speed
t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0
H = Vy t + 1/2 g t^2
H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m