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:
2(maximum), -2(minimum), -2(maximum).
Explanation:
V(t)= 2πcos πt--------------------------------------------------------------------------------(1).
Therefore, there is a need to integrate v(t) to get S(t).
S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).
Applying the condition given, we have s(0)= 0.
S(0)= 2sin ×π(0) + C.
Which means that; 0+C= 0. That is; C=0.
S(t)= 2 sin πt.
The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.
Take note that; sin(π/2) = sin(5π/2) = 1 .
Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.
Therefore, we have that 2 maximum; -2 minimum and -2 maximum.
