Test: Projectile Motion & Uniform Circular Motion (May 20) - JEE MCQ

Horizontal velocity of a projectile is not affected by gravity.

At the highest point, velocity is acting horizontally and acceleration ( = acceleration due to gravity) is acting vertically downwards. Therefore, at the highest point the angle between velocity and acceleration is 90^∘ .

Letbe the initial velocities of the two particles and θ_1​ and θ₂​ be their angles of projection with the horizontal.
The velocities of the two particles after time t are,

Time to reach maximum height = t_m
Time to reach back to ground = t_m
Total time of flight, T_f = t_m + t_m = 2t_m

Here, v  = 400ms⁻¹,

r = 0.6 km = 0.6 × 10³m,

g = 10ms⁻²
Centripetal acceleration,

= 26.7 × 10ms⁻²
In the units of (g = 10ms⁻²), the centripetal acceleration is 26.7.

In Circular motion, the Position vector is always outward and passes through the center. whereas the and centripetal acceleration is always toward the center. 
∴ Both is the opposite direction.

The kinematic equations for uniform acceleration do not apply in case of uniform circular motion because in this case the magnitude of acceleration is constant but its direction is changing.

Since the initial position coincides with the final position. So, the net displacement of the cyclist = zero

Given here,
Speed of the cyclist, v = 27 km/h = 7.5m/s
Radius of the circular tun on the road, r = 80 m
Therefore the centripetal acceleration of the cyclist at the moment when velocity is 7.5 m/s is,

When brakes are applied, the speed is decreased at the rate of 0.5m/s every second.

Therefore the tangential acceleration is,
a_r = - 0.5 m/s²
Therefore the total acceleration of the cyclist is,