In horizontal projectile motion, the initial vertical velocity is:
C
Equals horizontal velocity
D
Equals acceleration due to gravity
Solution:
In horizontal projectile motion, the object is launched horizontally, meaning the initial vertical velocity (v₀y) is zero. The object only has initial horizontal velocity.
The horizontal velocity of a projectile during flight:
D
First increases then decreases
Solution:
In projectile motion, there is no horizontal acceleration (ignoring air resistance). Therefore, the horizontal velocity remains constant throughout the flight.
A projectile launched horizontally will hit the ground at the same time as an object:
A
Thrown upward from the same height
B
Dropped from the same height
C
Thrown at an angle from the same height
Solution:
Since both objects start with zero vertical velocity and fall under the same gravitational acceleration from the same height, they will take the same time to reach the ground.
If a ball is thrown horizontally with velocity 15 m/s from height 45 m, find the horizontal distance traveled: (g = 10 m/s²)
Solution:
First find time: h = ½gt² → 45 = ½ × 10 × t² → t = 3 s
Horizontal distance = horizontal velocity × time = 15 × 3 = 45 m
In horizontal projectile motion, the acceleration in the horizontal direction is:
Solution:
In projectile motion (ignoring air resistance), there is no force acting horizontally, so the horizontal acceleration is zero.
If the height of projection is quadrupled, the time of flight becomes:
Solution:
Since t = √(2h/g), if height becomes 4h, then t becomes √(2×4h/g) = 2√(2h/g) = 2t. The time doubles.
A ball projected horizontally travels 40 m horizontally while falling 20 m vertically. Find the initial horizontal velocity: (g = 10 m/s²)
Solution:
First find time: h = ½gt² → 20 = ½ × 10 × t² → t = 2 s
Initial horizontal velocity = horizontal distance / time = 40 / 2 = 20 m/s
At the highest point of trajectory in horizontal projectile motion:
A
Both velocities are zero
B
Vertical velocity is zero, horizontal velocity is maximum
C
There is no highest point in horizontal projection
D
Horizontal velocity is zero
Solution:
In horizontal projectile motion, the object starts at its highest point and falls down. There is no point higher than the starting point, so there's no "highest point" during flight.
A projectile is launched horizontally from height h. If air resistance is neglected, the vertical distance fallen in the first second is:
C
Depends on horizontal velocity
Solution:
Using s = ut + ½at² with u = 0, a = g = 10 m/s², t = 1 s:
s = 0 + ½ × 10 × 1² = 5 m
Two balls are released simultaneously: one dropped vertically, another projected horizontally from the same height. Which hits the ground first?
C
Both hit simultaneously
D
Depends on horizontal velocity
Solution:
Both balls have the same initial vertical velocity (zero) and same vertical acceleration (g). Therefore, they take the same time to fall and hit the ground simultaneously.
A projectile launched horizontally from height 125 m hits the ground with vertical velocity 50 m/s. Find g:
Solution:
Using v² = u² + 2as with u = 0, v = 50 m/s, s = 125 m:
50² = 0² + 2 × g × 125
2500 = 250g
g = 10 m/s²
The equation of trajectory for horizontal projectile motion launched from height h₀ is:
Solution:
From kinematic equations: x = v₀t, y = h₀ - ½gt²
Eliminating t: t = x/v₀, so y = h₀ - ½g(x/v₀)² = h₀ - (g/2v₀²)x²
A ball is projected horizontally with velocity 25 m/s from a cliff. After 4 seconds, find its vertical velocity: (g = 10 m/s²)
Solution:
Vertical velocity after time t: v = u + gt = 0 + 10 × 4 = 40 m/s
